{"id":335,"date":"2017-08-28T04:30:55","date_gmt":"2017-08-28T04:30:55","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/mathematics\/?page_id=335"},"modified":"2017-09-22T19:01:26","modified_gmt":"2017-09-22T19:01:26","slug":"integral-calculus","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/mathematics\/integral-calculus\/","title":{"rendered":"Integral Calculus"},"content":{"rendered":"
\n
\u2b05 Previous Lesson<\/a>\u00a0Workshop Index<\/a>\u00a0Next\u00a0Lesson \u27a1<\/a><\/div>\n

<\/p>\n

Integral Calculus<\/h1>\n

Objective<\/h4>\n

In this lesson, you will define and interpret the meaning of definite and indefinite integrals, use integral calculus\u00a0in pure and applied problem-solving applications, learn several techniques of integration of functions, and learn\u00a0techniques to approximate the value of definite and improper integrals.<\/p>\n

Previously Covered:<\/h4>\n
    \n
  • The derivative<\/strong><\/em> of a function \u00a0is expressed as .<\/li>\n
  • The derivative can be\u00a0interpreted as the slope of the tangent line<\/em><\/strong> of at ,\u00a0or the instantaneous rate of change<\/strong><\/em> of at\u00a0.<\/li>\n
  • The first\u00a0derivative<\/em><\/strong> is useful in calculations involving the relative maxima and\u00a0minima<\/strong><\/em> of\u00a0functions, the intervals of increase and decrease<\/em><\/strong> of a function, the roots\u00a0<\/em><\/strong>of functions, optimization <\/em><\/strong>problems, and related\u00a0rate<\/em><\/strong> problems.<\/li>\n
  • The second\u00a0derivative<\/em><\/strong> of a function is helpful in determining concavity<\/em>,<\/strong> type\u00a0of relative\u00a0extrema<\/em><\/strong> within an interval, and inflection\u00a0points<\/em><\/strong>.<\/li>\n
  • The chain rule<\/strong><\/em>, implicit\u00a0differentiation<\/strong><\/em>, and parametric differentiation<\/em><\/strong> are\u00a0important principles in finding derivatives of composite\u00a0functions<\/em><\/strong> and implicit equations<\/em><\/strong>.<\/li>\n
  • L\u2019H\u00f4pital\u2019s rule<\/strong><\/em> is used to determine the limits of functions\u00a0of the indeterminate\u00a0form .<\/li>\n<\/ul>\n
    \n

    Differential Calculus II<\/h3>\n

    What is the meaning of an integral?<\/p>\n

    Most textbooks do not begin their discussion of integrals with\u00a0a formal definition, as they do with limits and derivatives of\u00a0functions. They typically begin by conceptualizing an indefinite\u00a0integral as the inverse of a derivative (or antiderivative<\/em>),\u00a0and a definite integral as the area under a curve<\/em>.<\/p>\n

    Suppose we have a function\u00a0\u00a0defined on\u00a0,\u00a0and a function\u00a0\u00a0which is continuous on\u00a0,\u00a0differentiable on\u00a0,\u00a0and\u00a0\u00a0for all\u00a0\u00a0in\u00a0.\u00a0Then,\u00a0\u00a0is known as the antiderivative<\/abbr> or indefinite\u00a0integral<\/abbr> of\u00a0.\u00a0This is denoted\u00a0\u00a0and is read as, \u201c\u00a0is the integral of\u00a0\u00a0with respect to\u00a0.\u201d\u00a0Using some of the known differentiation formulas from the previous\u00a0module, let\u2019s work out a couple of examples.<\/p>\n

    Find the antiderivative of\u00a0.<\/p>\n

    Recall that the derivative of\u00a0\u00a0is\u00a0.\u00a0Therefore, the derivative of\u00a0\u00a0is\u00a0,\u00a0which is the value of the function inside the integral (known as\u00a0the integrand<\/abbr>).\u00a0From this, we can see that\u00a0\u00a0is a solution. However, so is\u00a0\u00a0In fact,\u00a0\u00a0where\u00a0\u00a0is any constant.<\/p>\n

    What would be the answer to this antiderivative if the\u00a0coefficient\u00a0\u00a0were not there? Prove to yourself that the answer would be\u00a0.\u00a0From this, you should be able to derive the following theorem.<\/p>\n

    \n

    Important Tidbit<\/h4>\n

    <\/p>\n<\/div>\n

    A simple approach is to just take the derivative of the right-hand side. You will see that you get the integrand of the\u00a0left-hand side.<\/p>\n

    Before we work on the second example, let\u2019s look at a new\u00a0theorem.<\/p>\n

    \n

    Important Tidbit<\/h4>\n

    If and are\u00a0functions that have antiderivatives in the same interval, and if are constants, then<\/p>\n

    .<\/p>\n<\/div>\n

    Integral Calculus<\/h3>\n

    Let’s try the second example. Calculate\u00a0.<\/p>\n

    Using the above theorem, we rewrite the integral.<\/p>\n

    <\/p>\n

    To solve for\u00a0\u00a0and\u00a0,\u00a0recall that\u00a0\u00a0and\u00a0.\u00a0Therefore,\u00a0\u00a0and\u00a0,\u00a0and the solution to the indefinite integral (never forget the\u00a0)\u00a0is\u00a0.<\/p>\n

    The antiderivative, or indefinite integral, is a function, with\u00a0infinite possible solutions. Our second conceptualization of the\u00a0integral is the area under a curve<\/strong>, which is the\u00a0definite\u00a0integral<\/abbr> of the function. Unlike indefinite integrals, a\u00a0definite integral is a number rather than a function.<\/p>\n

    Let\u2019s look at the function\u00a0\u00a0over the interval\u00a0.\u00a0Suppose we wish to determine the area bounded by the curve\u00a0and the <\/span>x<\/span><\/em>-axis,\u00a0as depicted below. <\/span><\/p>\n

    \"Area<\/p>\n

    We can estimate the area under the curve by drawing rectangles\u00a0over fixed intervals (in this case, intervals of\u00a0),\u00a0with either the left corner or right corner of the rectangle\u2019s\u00a0upper width touching the curve, as depicted below.<\/p>\n\n\n\n
    <\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    The area inside the rectangles is easy to compute. The area in the first graph is\u00a0,\u00a0while\u00a0the area in the second graph is\u00a0.\u00a0We know the actual area\u00a0\u00a0under the curve is between these two values, .22 and .47. We can\u00a0narrow the range by taking smaller intervals, such as every\u00a0.<\/p>\n

    We then compute the new areas\u00a0\u00a0and\u00a0.<\/p>\n


    \nand<\/p>\n

    <\/p>\n

    <\/p>\n

    We still have not arrived at a good approximation of the area,\u00a0but the interval width has narrowed from \u00a0to\u00a0.\u00a0By continuing to narrow the intervals, both calculations will\u00a0approximate the true area under the curve (which is\u00a0)\u00a0more and more closely.<\/p>\n

    This method of measuring the area as a series of rectangles is\u00a0known as a Riemann\u00a0sum<\/abbr>. More specifically, the first graph is an example of a\u00a0left Riemann sum<\/em>, and the second graph is a right\u00a0Riemann sum<\/em>. The intervals,\u00a0\u00a0, for the two calculations are examples of partitions<\/strong>.\u00a0In the first calculation, the right partition contains\u00a0\u00a0subintervals<\/strong> of equal length. In the second\u00a0calculation, the right partition contains\u00a0\u00a0subintervals. For a function\u00a0,\u00a0we represent a Riemann sum using the notation\u00a0,\u00a0where\u00a0\u00a0represents the\u00a0\u00a0partition.<\/p>\n

    As the subintervals become smaller,\u00a0,\u00a0and the Riemann sum approaches the true area. In the closed\u00a0interval\u00a0,\u00a0we define the definite integral of the function\u00a0\u00a0as:<\/p>\n

    <\/p>\n

    We refer to\u00a0\u00a0as the lower limit <\/strong>and\u00a0\u00a0as the upper limit.<\/strong><\/p>\n

    To be precise, the area under the curve\u00a0\u00a0in the interval\u00a0\u00a0is not\u00a0\u00a0if the value of\u00a0\u00a0is negative in any part of the open interval. For example, look at\u00a0the area under the curve\u00a0\u00a0on the interval\u00a0.<\/p>\n

    \"<\/p>\n

    As we will see, if we were to evaluate the definite integral in\u00a0the interval\u00a0,\u00a0we would find\u00a0,\u00a0since the shaded area below the x<\/em>-axis cancels out the\u00a0symmetrically shaded area above the x<\/em>-axis. To rectify\u00a0this, we define the area as\u00a0.\u00a0In this example, we split the integral into two portions and find\u00a0the sum of the positive values. Thus,\u00a0.<\/p>\n

    What is the Fundamental Theorem of Calculus?<\/h3>\n

    In order to complete our understanding\u00a0of\u00a0definite integrals, we state and prove the Fundamental\u00a0Theorem of Calculus<\/abbr>.<\/p>\n

    \n

    The Fundamental Theorem of Calculus<\/h4>\n

    Let be a function continuous on and is any number in . <\/span><\/em><\/p>\n

    (i) If we define , then . <\/span><\/em><\/p>\n

    (ii) If we define , then . <\/span><\/em><\/p>\n<\/div>\n

    Proof of:
    \n<\/strong><\/p>\n

    Picture\u00a0\u00a0as the area under the curve\u00a0\u00a0from\u00a0\u00a0to\u00a0.\u00a0Then\u00a0\u00a0is the area from\u00a0\u00a0to\u00a0,\u00a0and\u00a0\u00a0is the area from\u00a0\u00a0to\u00a0.\u00a0If\u00a0\u00a0is small, the area from\u00a0\u00a0to\u00a0\u00a0should be small. Also, since\u00a0\u00a0is continuous,\u00a0.\u00a0That area should then approximate a rectangle of height\u00a0\u00a0and width\u00a0.\u00a0Therefore,<\/p>\n

    , or\u00a0.\u00a0As\u00a0,\u00a0this approximation becomes exact, so\u00a0.<\/p>\n

    Proof of ():
    \n<\/strong><\/p>\n

    Given\u00a0,\u00a0and using (i), it follows that\u00a0,\u00a0which means\u00a0\u00a0,or\u00a0.\u00a0Therefore,\u00a0\u00a0,and\u00a0,\u00a0so\u00a0.\u00a0The second integral is clearly 0 (imagine the area of a rectangle\u00a0with width 0). Therefore,\u00a0.<\/p>\n

    Part (ii) of the theorem states that the value of the definite\u00a0integral on the interval\u00a0\u00a0is simply the difference of the antiderivative at the endpoints.\u00a0Our shorthand is\u00a0.\u00a0Let\u2019s try a couple of examples solving definite integrals.<\/p>\n

    Find the actual areas under the curves on the given intervals:<\/p>\n

    (A) area under\u00a0\u00a0over the interval\u00a0<\/p>\n

    (B) area under\u00a0\u00a0over the interval\u00a0<\/p>\n

    (A)\u00a0\u00a0everywhere in the given interval, so\u00a0.<\/p>\n

    (B)\u00a0\u00a0on the interval\u00a0,\u00a0and\u00a0\u00a0on the interval\u00a0,\u00a0so<\/p>\n

    <\/p>\n

    Integral Calculus<\/h3>\n

    A brick is dropped from the top of a building that is 500\u00a0meters tall. Its velocity after\u00a0\u00a0seconds is given by\u00a0\u00a0meters per second. How long before the brick hits the ground?<\/p>\n

    You may recall that we found\u00a0\u00a0by taking the first derivative of the position\u00a0;\u00a0that is,\u00a0.\u00a0Here, we are given\u00a0\u00a0and need to solve for\u00a0,\u00a0which is the antiderivative of the velocity function. From part\u00a0(ii) of the Fundamental Theorem of Calculus, we know that\u00a0.\u00a0In this example, a<\/em> represents the time corresponding to the\u00a0start of the fall and b<\/em> represents the time corresponding\u00a0to the end of the fall, the time we are looking for. So we will\u00a0set\u00a0\u00a0seconds,\u00a0\u00a0seconds, and solve for\u00a0.\u00a0We know\u00a0\u00a0m and\u00a0.\u00a0Therefore,\u00a0.\u00a0Also, the total change in the brick\u2019s position is simply its\u00a0final position minus its initial position, or\u00a0\u00a0meters. Therefore,\u00a0,\u00a0or\u00a0\u00a0seconds ().<\/p>\n

    Before we proceed with the next example, let\u2019s explore a\u00a0notational detail which may seem strange at first. Recall that the\u00a0expression\u00a0\u00a0can be read as, \u201cG<\/em> of x<\/em> is the integral of f\u00a0<\/em>of x<\/em>, with respect to x.<\/em>\u201d What if f<\/em>(x<\/em>)\u00a0is a constant function? If f<\/em>(x<\/em>) = 5, then\u00a0\u00a0=\u00a0.\u00a0This can be read as, \u201cG<\/em> of x<\/em> is 5 times the\u00a0integral of 1, with respect to x<\/em>.\u201d We now know this\u00a0is 5x<\/em> + C<\/em>, where C<\/em> is some constant of\u00a0integration. Typically, this last term is written in the\u00a0simplified form .\u00a0If the integrand symbol followed\u00a0by the <\/span>dx\u00a0<\/span><\/em>with nothing in the middle seems strange, imagine a constant of 1\u00a0sitting between the integrand symbol and the dx. <\/em><\/span><\/p>\n

    Now let\u2019s try the example.<\/h4>\n

    The marginal revenue (rate of change of revenue received per\u00a0unit change in quantity sold) received by a retailer is given by\u00a0,\u00a0where\u00a0\u00a0is the quantity of units sold. Compute the total revenue ()\u00a0as a function of\u00a0.<\/p>\n

    According to our definition of marginal revenue,\u00a0.\u00a0Therefore, we compute\u00a0\u00a0as the antiderivative of\u00a0.<\/p>\n

    <\/p>\n

    To find a complete solution to the problem, we must now solve\u00a0for C<\/em> by checking to see what value of C<\/em> will make\u00a0our revenue function accurately reflect the conditions set forth\u00a0in the original problem statement. If we sell a quantity of 0, we\u00a0make no revenue.<\/p>\n

    ,
    \nor
    \n.<\/p>\n

    Our final solution is\u00a0.<\/p>\n

    How do we find the area bounded by two curves?<\/h3>\n

    We\u2019ve developed a formula for the area under the curve\u00a0\u00a0in\u00a0:\u00a0.\u00a0More specifically, this formula describes the area under the curve\u00a0\u00a0bounded by the lines\u00a0,\u00a0and the <\/span>x<\/span><\/em>-axis.\u00a0Let\u2019s now consider the <\/span>area\u00a0bounded by two curves, <\/span><\/strong>\u00a0and\u00a0.\u00a0Let\u00a0,\u00a0and\u00a0,\u00a0and look at the area under each curve on\u00a0.
    \n<\/span><\/p>\n\n\n\n\n
    \"Area<\/td>\n\"Area<\/td>\n<\/tr>\n
    <\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    The curves intersect at\u00a0\u00a0and\u00a0,\u00a0and their individual areas are represented by\u00a0\u00a0and\u00a0.<\/p>\n

    Now let\u2019s look at both graphs and the area between them\u00a0on\u00a0.<\/p>\n

    \"<\/p>\n

    The area between the curves\u00a0is the difference between the areas under each separate curve\u00a0bounded by the <\/span>x<\/span><\/em>-axis.\u00a0On\u00a0,\u00a0,\u00a0and the area is represented by <\/span><\/p>\n

    <\/p>\n

    On\u00a0,\u00a0,\u00a0and the area is<\/p>\n

    <\/p>\n

    Therefore, to find the area\u00a0\u00a0on\u00a0\u00a0we need to solve the definite integrals.<\/p>\n

    <\/p>\n

    \n

    Important Tidbit<\/h4>\n

    The area between two curves and\u00a0,\u00a0between and is .<\/p>\n<\/div>\n

    In some applications, it may be easier to compute the area of\u00a0the region bounded by two curves by looking at horizontal elements\u00a0rather than vertical rectangles. For example, look at the region\u00a0bounded by the parabola\u00a0\u00a0and the line\u00a0.\u00a0The area of the region with a vertical rectangle is shown below.<\/p>\n

    \"Area<\/p>\n

    Note that the interval is on\u00a0,\u00a0and the line becomes the lower bound on\u00a0.\u00a0To compute the area this way, first, we need to solve the first\u00a0equation for\u00a0.<\/p>\n

    <\/p>\n

    These separate equations represent the upper and lower halves\u00a0of the parabola.<\/p>\n

    We solve for the total area in two parts. On\u00a0,\u00a0the integrand is the difference between the upper and lower halves\u00a0of the parabola.<\/p>\n

    <\/p>\n

    On\u00a0,\u00a0the integrand is the difference between the upper half of the\u00a0parabola and the line.<\/p>\n

    <\/p>\n

    The total area is the sum .<\/p>\n

    Let\u2019s look at this problem again, but this time, we will\u00a0compute the area by taking horizontal rectangular elements. Note\u00a0that the curves intersect at\u00a0\u00a0and\u00a0.<\/p>\n

    First, we need to solve both equations for\u00a0.\u00a0We get\u00a0\u00a0and\u00a0.\u00a0The interval of integration is on the y-<\/em>range\u00a0,\u00a0and the heights of the rectangles are\u00a0.\u00a0Therefore, we can compute the total area between the curves with a\u00a0single integral.<\/p>\n

    <\/p>\n

    Clearly, this method is easier than the first.<\/p>\n

    How do we find the volume of a solid of revolution?<\/h3>\n

    Another application of definite integrals is calculating the\u00a0volume of a solid of revolution<\/strong>. This is the\u00a0solid created by revolving a portion of a curve about a line, such\u00a0as the x<\/em>– or y<\/em>-axis. Let\u2019s take\u00a0\u00a0on\u00a0\u00a0and see what happens if we revolve that portion of the curve about\u00a0the x<\/em>-axis and y<\/em>-axis.<\/p>\n\n\n\n\n
    \"Curve<\/td>\n\"Curve<\/td>\n<\/tr>\n
    x-axis<\/td>\ny-axis<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    For the first graph, imagine a rectangular sliver of width\u00a0\u00a0as it revolves around the x<\/em>-axis. It forms a circular disk\u00a0of volume\u00a0.\u00a0We can approximate the volume as the sum of rectangles along the\u00a0interval, just as we have been approximating the areas, but we\u00a0will skip directly to what happens as\u00a0.\u00a0As\u00a0,\u00a0the sum of rectangles becomes the definite integral on the\u00a0interval\u00a0,\u00a0and the volume becomes\u00a0.<\/p>\n

    For example, solve\u00a0.<\/p>\n

    Suppose we wish to revolve this same function on the same\u00a0interval as in the second graph about the y<\/em>-axis. First, we\u00a0need to solve for\u00a0.\u00a0(In this case,\u00a0,\u00a0as we are only looking at\u00a0.)\u00a0Each sliver forms a disk of volume\u00a0,\u00a0and we integrate on the\u00a0y<\/span><\/em>-interval\u00a0<\/span>.\u00a0The volume of the solid revolved around the y<\/em>-axis is\u00a0.<\/p>\n

    Finally, the volume of the solid of revolution bounded by two\u00a0curves\u00a0\u00a0and\u00a0\u00a0is found by using reasoning similar to that which we have used\u00a0with areas. If both functions are continuous and\u00a0\u00a0over\u00a0,\u00a0then the volume of the solid of revolution of the region bounded\u00a0by\u00a0\u00a0and\u00a0\u00a0about the x<\/em>-axis is<\/p>\n

    <\/p>\n

    Integral Calculus<\/h3>\n

    Another computation it is\u00a0useful to be able to find is the <\/span>area\u00a0of surface of revolution<\/span><\/strong>.\u00a0The area of the surface of revolution obtained by rotating\u00a0\u00a0on\u00a0\u00a0about the x<\/em>-axis is <\/span><\/p>\n

    <\/p>\n

    The area of the surface of\u00a0revolution obtained by rotating\u00a0\u00a0on the <\/span>y<\/span><\/em>-interval\u00a0\u00a0about the y<\/em>-axis is <\/span><\/p>\n

    <\/p>\n

    Another application of definite integrals is finding the length\u00a0of a curve<\/strong>, or arc length. You should recall that the\u00a0distance between point\u00a0\u00a0and point\u00a0\u00a0is\u00a0.\u00a0We can approximate the arc length of a curve continuous on\u00a0\u00a0as the sum of straight lines between points\u00a0\u00a0and\u00a0.\u00a0Each small segment has a length\u00a0.\u00a0As the length intervals become infinitesimally small, the exact\u00a0length can be computed as the definite integral on\u00a0.\u00a0The value\u00a0\u00a0approaches\u00a0\u00a0(the proof of this utilizes the mean value theorem) and\u00a0.\u00a0As long as\u00a0\u00a0and\u00a0\u00a0are continuous on\u00a0,\u00a0the length of the curve from\u00a0\u00a0to\u00a0\u00a0is given by<\/p>\n

    <\/p>\n

    If we want to find the length of the curve\u00a0\u00a0from\u00a0\u00a0to\u00a0\u00a0and\u00a0\u00a0and\u00a0\u00a0are continuous on\u00a0,\u00a0the length of the curve is<\/p>\n

    <\/p>\n

    The final application of definite integrals we will examine is\u00a0the work done by a force acting on an object. If the object is\u00a0acted on by a force\u00a0\u00a0and is displaced along the x<\/em>-axis from point\u00a0\u00a0to point\u00a0,\u00a0the work<\/strong> done by the force is\u00a0.\u00a0Work is usually measured in foot-pounds, or, if the force is in\u00a0newtons and the displacement is in meters, in joules. Sometimes,\u00a0force is measured in dynes and displacement in centimeters. Then,\u00a0work is measured in dyne-cm, or ergs.<\/p>\n

    Let’s try an example. Consider the curve\u00a0\u00a0from\u00a0\u00a0to\u00a0.\u00a0Compute the following:<\/p>\n

    (A) the area under the\u00a0curve bounded by the <\/span>x-<\/span><\/em>axis,\u00a0using vertical rectangles; <\/span><\/p>\n

    (B) the area under the\u00a0curve bounded by the <\/span>x-<\/span><\/em>axis,\u00a0using horizontal rectangular elements; <\/span><\/p>\n

    (C) the volume of the solid\u00a0of revolution about the <\/span>x-<\/span><\/em>axis;\u00a0and <\/span><\/p>\n

    (D) the volume of the solid\u00a0of revolution about the <\/span>y-<\/span><\/em>axis.
    \n<\/span><\/p>\n

    (A)\u00a0<\/p>\n

    (B) Solve\u00a0\u00a0for x. \u00a0<\/em>.\u00a0At\u00a0,\u00a0and at\u00a0.\u00a0The height of each rectangle is\u00a0,\u00a0and its width is\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    (C)\u00a0<\/p>\n

    (D)\u00a0<\/p>\n

    Let\u2019s try a second example. For the line\u00a0,\u00a0find:<\/p>\n

    (A) the length of arc from\u00a0\u00a0to\u00a0;\u00a0and<\/p>\n

    (B) the area of surface of revolution about the x<\/em>-axis\u00a0over the same interval.<\/p>\n

    (A) At\u00a0,\u00a0;\u00a0at\u00a0,\u00a0.\u00a0Also,\u00a0,\u00a0so\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    Since the curve in question is simply a line, we can\u00a0verify this solution by using the distance formula to find the\u00a0distance between the points (9, 7) and (18, 10). Indeed, the\u00a0distance formula gives us the same answer for the length of the\u00a0arc from\u00a0\u00a0to\u00a0.<\/p>\n

    <\/p>\n

    What is integration by substitution?<\/h3>\n

    So far, we have solved definite and indefinite integrals where\u00a0the integrand is of the form\u00a0,\u00a0,\u00a0and\u00a0.\u00a0This section examines integrals of various functions, and\u00a0techniques for solving integrals of various forms.<\/p>\n

    The first method we will examine is integration by\u00a0substitution<\/strong>. We know how to solve for integrals in the\u00a0form\u00a0\u00a0().\u00a0For example, we know how to solve\u00a0.\u00a0This method does not work if the power of\u00a0.\u00a0However, we can use substitution to solve a similar integral:\u00a0.\u00a0Let\u00a0,\u00a0so\u00a0\u00a0and substitute\u00a0\u00a0for\u00a0.<\/p>\n

    <\/p>\n

    Thus, if we have an integral whose integrand can be placed in\u00a0the form\u00a0\u00a0,\u00a0we can use the chain rule and substitution to transform the\u00a0integral to the form\u00a0.\u00a0Then, it can be easily integrated.<\/p>\n

    For the case\u00a0,\u00a0we can determine\u00a0.\u00a0Recall from differential calculus that\u00a0.\u00a0But\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    Let’s try a few examples. Compute\u00a0.<\/p>\n

    Let\u00a0.\u00a0Then\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    Compute\u00a0.\u00a0Let\u00a0,\u00a0and\u00a0.\u00a0Therefore<\/p>\n

    <\/p>\n

    What is integration by parts?<\/p>\n

    Another integration technique is known as integration\u00a0by parts<\/strong>. If an integral is in the form\u00a0,\u00a0but can be solved in the form\u00a0,\u00a0use the following.<\/p>\n

    <\/p>\n

    This formula is derived from the product rule in differential\u00a0calculus. Recall that\u00a0.\u00a0When solving for\u00a0,\u00a0we find that\u00a0.\u00a0Finally, integrate all three terms to arrive at the formula. The\u00a0goal of integration by parts is to transform one integral into\u00a0another integral that is \u201csolvable.\u201d After deciding\u00a0what to call\u00a0\u00a0and\u00a0\u00a0and solving for\u00a0\u00a0and\u00a0,\u00a0one of three things should happen:<\/p>\n

    (1)\u00a0is as hard or harder to solve than\u00a0,\u00a0and the process is abandoned.<\/p>\n

    (2)\u00a0\u00a0is easy to solve, and the formula is straightforward.<\/p>\n

    (3) Solving for\u00a0\u00a0leads to a solution with\u00a0,\u00a0which is solved by algebra.<\/p>\n

    The following examples illustrate each of these scenarios.<\/p>\n

    Compute\u00a0\u00a0by setting\u00a0\u00a0and\u00a0.<\/p>\n

    We find that\u00a0\u00a0Therefore,\u00a0\u00a0However, this new integral is even more difficult to solve than\u00a0the original, so we abandon this process.<\/p>\n

    Compute\u00a0\u00a0by setting\u00a0\u00a0and\u00a0.<\/p>\n

    We find that\u00a0.\u00a0Therefore,\u00a0.\u00a0We can easily solve this new integral by substitution. Let\u00a0,\u00a0so\u00a0.\u00a0The integral is\u00a0.\u00a0So,<\/p>\n

    <\/p>\n

    Integral Calculus<\/h3>\n

    These first two examples illustrate an important point about\u00a0integration problems and the methods used. For most of our\u00a0mathematical career, mathematics is generally algorithmic. As long\u00a0as rules and procedures are memorized and practiced, a solution\u00a0can usually be reached through careful perseverance. This is not\u00a0the case when working with integrals. In fact, the integrals of\u00a0most functions, if they are randomly drawn from the \u201cuniverse\u00a0of functions,\u201d are not solvable manually by any of our\u00a0methods. Even if they are solvable, not every possible method will\u00a0get us to the solution. Don\u2019t let this frustrate you, and\u00a0try to remember that with practice, you will begin to recognize\u00a0patterns in the structure of the functions you are integrating\u00a0which give clues as to possible methods that might be worth\u00a0trying.<\/p>\n

    Compute\u00a0.<\/p>\n

    First, let\u00a0.\u00a0Set\u00a0,\u00a0and\u00a0.\u00a0We find that\u00a0.\u00a0Therefore,\u00a0\u00a0To solve this new integral, perform integration by parts. Set\u00a0\u00a0and\u00a0.\u00a0It is then true that\u00a0,\u00a0so\u00a0\u00a0.\u00a0Now, we have\u00a0\u00a0If we apply simple algebra, we find that\u00a0\u00a0or\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    What is trigonometric substitution?<\/h4>\n

    If the integrand is in the form\u00a0,\u00a0we can solve the integral by trigonometric substitution<\/strong>.\u00a0Recall the trigonometric identities\u00a0\u00a0and\u00a0.\u00a0If we set\u00a0,\u00a0the term\u00a0\u00a0simplifies to\u00a0\u00a0 If we set\u00a0,\u00a0the term\u00a0\u00a0simplifies to\u00a0\u00a0.\u00a0And if we set\u00a0,\u00a0the term\u00a0\u00a0simplifies to\u00a0<\/p>\n

    Integral Calculus<\/h3>\n

    Trigonometric substitution is often easier with definite\u00a0integrals than with indefinite integrals. With definite integrals,\u00a0we can leave the answer in terms of\u00a0\u00a0and change the upper and lower limit of the integral into terms of\u00a0\u00a0rather than\u00a0,\u00a0where appropriate. With indefinite integrals, we must change\u00a0\u00a0to\u00a0,\u00a0which can be tricky.\u00a0Let\u2019s try an example. Compute\u00a0.\u00a0Hint:\u00a0<\/em>.<\/p>\n

    Let\u00a0,\u00a0so\u00a0.\u00a0Since the 0 and 1 correspond to x<\/em>-values, not \u03b8<\/em>-values,\u00a0we must change the limits of integration. Since\u00a0,\u00a0and we are now integrating with respect to <\/span>\u03b8<\/span><\/em>,\u00a0we look for a \u03b8 <\/em>value which will make sin\u00a0<\/span>\u03b8 <\/span><\/em>= 0. This\u00a0requirement is met when \u03b8<\/em> = 0. The new upper limit,\u00a0which corresponds to\u00a0,\u00a0is satisfied when\u00a0<\/span>,\u00a0or\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    The next integration technique is useful when the integrand can\u00a0be expressed as a rational\u00a0function<\/abbr>, that is, the quotient of two polynomial\u00a0functions\u00a0.\u00a0This method involves breaking up\u00a0\u00a0into the sum of partial fractions<\/strong>. For example,\u00a0if the integrand is\u00a0,\u00a0create partial fractions by setting\u00a0.\u00a0By rewriting each term on the right so that the denominators are\u00a0identical to the denominator on the left and reducing, we have\u00a0\u00a0Factoring and rearranging terms gives\u00a0\u00a0We now have 3 equations and 3 unknowns\u00a0.\u00a0The solution you should get is\u00a0.\u00a0We can use these terms to rewrite the integrand as the sum of\u00a0three simple terms and integrate each one separately.<\/p>\n

    Here is another example. Compute\u00a0.<\/p>\n

    Using partial fractions, we rewrite this integral as<\/p>\n

    <\/p>\n

    When the denominator has factors in the form\u00a0\u00a0for\u00a0,\u00a0all factors of\u00a0\u00a0must be included in partial fractions. For example, the quotient\u00a0\u00a0is broken up into partial fractions\u00a0.<\/p>\n

    Integral Calculus<\/h3>\n

    Another technique involving rational functions is completing\u00a0the square<\/strong>. This technique is illustrated in the\u00a0following example.<\/p>\n

    Compute\u00a0\u00a0by completing the square.<\/p>\n

    <\/p>\n

    Set\u00a0,\u00a0so\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    We solve this integral by partial fractions:<\/p>\n

    .<\/p>\n

    Therefore,\u00a0<\/p>\n

    The solution is\u00a0,\u00a0so<\/p>\n

    <\/p>\n

    <\/p>\n

    The next form of function requires an understanding of inverse\u00a0trigonometric functions<\/strong>. For example, the inverse sine\u00a0function\u00a0\u00a0is satisfied if\u00a0.\u00a0We know\u00a0\u00a0is in the interval\u00a0.\u00a0In order for each value of x<\/em> in this interval to have a\u00a0unique\u00a0,\u00a0we restrict\u00a0\u00a0to the interval\u00a0.\u00a0The inverse cosine function\u00a0\u00a0exists if and only if\u00a0\u00a0on\u00a0.\u00a0The inverse tangent function\u00a0\u00a0exists if and only if\u00a0\u00a0on\u00a0.\u00a0The inverse secant function\u00a0<\/p>\n

    \n

    Important Tidbit<\/h4>\n

    From the formulas for the derivatives of the inverse trigonometric functions,\u00a0we obtain the following indefinite integrals.<\/p>\n

    <\/p>\n

    <\/p>\n<\/div>\n

    Techniques such as substitution and completing the square are\u00a0often required in order to factor the integrand into one of the\u00a0above forms.<\/p>\n

    Using the integration techniques introduced in this module\u00a0along with other techniques and tricks, we are capable of\u00a0integrating and, in the case of definite integrals, finding an\u00a0exact numerical solution. Unfortunately, exact solutions cannot be\u00a0found for all definite integrals, such as\u00a0\u00a0and\u00a0.\u00a0For these circumstances, we rely on methods of numerical or\u00a0approximate integration. We already examined a method of\u00a0approximating the value of a definite integral: the Riemann sum.\u00a0However, the example we looked at demonstrates that this is a\u00a0slow-converging approximation method. We will now examine two\u00a0numerical methods that provide better accuracy: the\u00a0<\/span>trapezoidal rule\u00a0<\/span><\/em>and <\/span>Simpson\u2019s\u00a0rule<\/span><\/em>. <\/span><\/p>\n

    With Riemann sums, the area of each subinterval is approximated\u00a0as a rectangle along the partition. The trapezoidal rule<\/abbr> more accurately defines each area as a trapezoid, which\u00a0better represents the change in the function. We illustrate this\u00a0point by referring back to our original function,\u00a0.\u00a0Compare the areas on the interval\u00a0\u00a0of a left Riemann sum, right Riemann sum, and trapezoid with the\u00a0exact area\u00a0.<\/p>\n\n\n\n\n
    <\/td>\n<\/td>\n<\/td>\n<\/tr>\n
    \n
    Area
    \nunder a curve 1<\/em><\/span><\/span><\/div>\n<\/td>\n
    		\n
    Area
    \nunder curve 2<\/em><\/span><\/span><\/div>\n<\/td>\n
    		\n
    Area
    \nunder curve 3<\/em><\/span><\/span><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    The area of the first region is\u00a0.\u00a0The area of the second region is \u00a0.\u00a0The area of the third region is .\u00a0The true area is .<\/p>\n

    Clearly, the trapezoid is closer to the true area than either\u00a0rectangle. (The trapezoidal area is actually the average of .)<\/p>\n

    If\u00a0\u00a0is continuous on\u00a0\u00a0and\u00a0\u00a0are regular partitions of\u00a0,\u00a0then the trapezoidal rule states:<\/p>\n

    <\/p>\n

    Integral Calculus<\/h3>\n

    Now use the trapezoidal rule to estimate the area under the\u00a0curve\u00a0\u00a0with\u00a0.<\/p>\n

    This is the same interval we examined previously. Recall that\u00a0we determined\u00a0\u00a0using Riemann sums. First, compute\u00a0.\u00a0Therefore,<\/p>\n

    <\/p>\n

    The trapezoid formula converges to the true area\u00a0\u00a0with 8 intervals more efficiently than either Riemann sum.<\/p>\n

    The trapezoidal rule, like Riemann sums, approximates the\u00a0definite integral as the sum of areas formed by straight lines\u00a0connecting two points. In contrast, Simpson\u2019s rule<\/abbr> computes the area bounded by parabolas connecting three points. This gives us a better approximation to the area under the\u00a0curve. Simpson\u2019s rule states:<\/p>\n

    <\/p>\n

    Note that Simpson\u2019s rule has\u00a0\u00a0subintervals while the trapezoid rule has\u00a0\u00a0subintervals.<\/p>\n

    For example, use Simpson\u2019s rule to compute\u00a0\u00a0with\u00a0.<\/p>\n

    <\/p>\n

    =\u00a0<\/p>\n

    The reason this value is the exact area is that Simpson\u2019s\u00a0rule assumes each partition is a parabola, and in this case,\u00a0\u00a0is an actual parabola.<\/p>\n

    We conclude our analysis of integral calculus with improper\u00a0integrals<\/strong>. This is a type of definite integral\u00a0\u00a0where either\u00a0,\u00a0or\u00a0\u00a0becomes infinite for some value of\u00a0\u00a0on\u00a0.<\/p>\n

    \n

    Important Tidbit<\/h4>\n

    Provided that is continuous on the\u00a0interval defined by the lower and upper limits of the integral, the following rules apply.<\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    for some real number <\/p>\n<\/div>\n

    If the limit exists and is finite, the improper integral is\u00a0convergent. <\/em>If the improper integral does not exist, it is\u00a0divergent<\/em>. The integral diverges if it blows up to\u00a0,\u00a0or if it has no defined limit. An example of the latter situation\u00a0is\u00a0.<\/p>\n

    Compute the area bounded by the\u00a0<\/span>x<\/span><\/em>-axis\u00a0on\u00a0\u00a0for the following curves: <\/span><\/p>\n

    (A);\u00a0and<\/p>\n

    (B).<\/p>\n

    (A).<\/p>\n

    (B).<\/p>\n

    This limit does not exist\u2014it blows up to\u00a0.\u00a0This improper integral is divergent.<\/p>\n

    Another type of improper integral is one that blows up at one\u00a0or more points on a finite interval. Such a point is represented\u00a0by a vertical asymptote. If\u00a0\u00a0has a vertical asymptote on\u00a0\u00a0at\u00a0,\u00a0and if\u00a0\u00a0exist for\u00a0\u00a0in\u00a0,\u00a0and for\u00a0\u00a0in\u00a0,\u00a0then<\/p>\n

    <\/p>\n

    Compute\u00a0.<\/p>\n

    The integrand has an infinite discontinuity at\u00a0,\u00a0as illustrated below<\/p>\n

    \"Integrand<\/p>\n

    Applying the above rule to this integral, with\u00a0,\u00a0we find<\/p>\n

    <\/p>\n

    Since neither of these limits exists, the integral is\u00a0divergent. Had we not recognized the discontinuity and attempted\u00a0to solve the definite integral, we would have arrived at the wrong\u00a0answer of\u00a0.\u00a0This answer makes no sense, since the integrand is always\u00a0positive.<\/p>\n<\/section>\n

    <\/p>\n

    \u2b05 Previous Lesson<\/a>\u00a0Workshop Index<\/a>\u00a0NextLesson \u27a1<\/a><\/div>\n

    Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    \u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next\u00a0Lesson \u27a1 Integral Calculus Objective In this lesson, you will define and interpret the meaning of definite and indefinite integrals, use integral calculus\u00a0in pure and applied problem-solving applications, learn several techniques of integration of functions, and learn\u00a0techniques to approximate the value of definite and improper integrals. Previously Covered: The derivative of a […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-335","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/335","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/comments?post=335"}],"version-history":[{"count":10,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/335\/revisions"}],"predecessor-version":[{"id":810,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/335\/revisions\/810"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/media?parent=335"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}