{"id":9350,"date":"2026-06-01T21:33:48","date_gmt":"2026-06-01T21:33:48","guid":{"rendered":"https:\/\/kapdec.com\/help\/?p=9350"},"modified":"2026-06-01T21:33:48","modified_gmt":"2026-06-01T21:33:48","slug":"value-of-function-by-definite-integral","status":"publish","type":"post","link":"https:\/\/kapdec.com\/help\/value-of-function-by-definite-integral\/","title":{"rendered":"Value Of Function By Definite Integral"},"content":{"rendered":"

Unit: <\/strong>Application of Integrations<\/strong><\/h2>\n

Chapter: <\/strong>Value of Function by Definite Integral<\/strong><\/h3>\n

Reference: – Differentiability & Continuity, Increasing & Decreasing functions, Curve sketching, Analysis of a function, Optimization problems, First & Second Derivative test, Related rates, Local extrema, Implicit differentiation & Applications<\/em>.<\/em><\/p>\n

 <\/p>\n

After studying this chapter, you should be able to:<\/strong><\/p>\n

    \n
  • Identifying the Function & the Interval.<\/li>\n
  • Evaluating Definite Integral<\/li>\n
  • Finding antiderivatives at the specified interval<\/li>\n<\/ul>\n

     <\/p>\n

    Introduction & Identification<\/u><\/strong><\/p>\n

     <\/p>\n

    The value of a function can be determined using a definite integral by evaluating the area under the curve of the function over a specific interval. The definite integral allows us to calculate the net area between the curve and the x-axis within the given interval.<\/p>\n

    Here's the process to find the value of a function using a definite integral: –<\/p>\n

      \n
    • Identify the function:<\/u><\/strong> Start by determining the function for which you want to find the value. For example, let's consider the function f(x).<\/li>\n<\/ul>\n

       <\/p>\n

        \n
      • Define the interval<\/u><\/strong>: Specify the interval over which you want to evaluate the function. This interval is usually denoted by [a, b], where 'a' is the lower limit and 'b' is the upper limit. This interval represents the range of x-values for which you want to calculate the value of the function.<\/li>\n<\/ul>\n

         <\/p>\n

          \n
        • Set up the definite integral<\/u><\/strong>: Write down the definite integral that represents the area under the curve of the function within the given interval. The definite integral is denoted by the symbol ∫ and includes the function, the variable of integration, and the interval. The general form of a definite integral is:<\/li>\n<\/ul>\n

          ∫[a, b] f(x) dx<\/p>\n

          Here, f(x) represents the function you want to evaluate, and dx indicates the variable of integration, which is typically x.<\/p>\n

           <\/p>\n

            \n
          • Evaluate the definite integral<\/u><\/strong>: Calculate the definite integral using integration techniques. This involves finding the antiderivative (or primitive) of the function f(x), denoted by F(x), and applying the Fundamental Theorem of Calculus. Then, substitute the upper limit (b) and the lower limit (a) into F(x) and subtract the result at the lower limit from the result at the upper limit:-<\/li>\n<\/ul>\n

            F(b) – F(a)<\/p>\n

            This computation provides the value of the function over the specified interval.<\/p>\n

             <\/p>\n

            Evaluating Definite Integral:<\/u><\/strong><\/p>\n

             <\/p>\n

            To evaluate  <\/em>\"\" by substitution, the steps could be as follows:<\/p>\n

            1. Consider the integral without limits and substitute, y <\/em>= f <\/em>(x<\/em>) or x <\/em>= g<\/em>(y<\/em>) to reduce<\/p>\n

            the given integral to a known form.<\/p>\n

            2. Integrate the new integrand for the new variable without mentioning<\/p>\n

            the constant of integration.<\/p>\n

            3. Resubstitute for the new variable and write the answer in terms of the original<\/p>\n

            variable.<\/p>\n

            4. Find the values of answers obtained in (3) at the given limits of integral and find<\/p>\n

            the difference of the values at the upper and lower limits.<\/p>\n

            Let us illustrate this by examples.<\/p>\n

            Example:<\/strong> Evaluate \"\"  dx.<\/p>\n

            Solution:<\/strong> Put t = x5<\/sup> + 1<\/p>\n

            dt = 5x4<\/sup> dx<\/p>\n

            Therefore, <\/p>\n

            \"\"  <\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

            \"\"  <\/p>\n

             \"\"<\/p>\n

             \"\"<\/p>\n

            Alternate:<\/strong><\/p>\n

            Let t = x5<\/sup> + 1<\/p>\n

            dt = 5x4<\/sup> dx.<\/p>\n

            When x = – 1, t = 0 and when x = 1, t = 2<\/p>\n

            Thus, as x varies from – 1 to 1, t varies from 0 to 2<\/p>\n

            Therefore,<\/p>\n

            \"\"  <\/p>\n

             \"\"<\/p>\n

             \"\"<\/p>\n

            \"\"<\/p>\n

            Example:<\/strong> Evaluate \"\" .<\/p>\n

            Solution: <\/strong>Let u = x2<\/sup> + 2<\/p>\n

            du = 2x dx<\/p>\n

            When x = -2, u = 6<\/p>\n

            When x= -1, u = 3<\/p>\n

            Therefore,<\/p>\n

            \"\"  <\/p>\n

            \"\"<\/p>\n

            \"\"<\/p>\n

            \"\"<\/p>\n

             \"\"<\/p>\n

                   \"\"             <\/p>\n

                                       (Root Mean Value of a Function)      <\/p>\n

             <\/p>\n

            Properties of a function by Definite Integral<\/u><\/strong>: –<\/strong><\/p>\n

             <\/p>\n

            Some of the basic yet important properties which prove fruitful while attempting questions on definite integral are listed below. These results can be proved easily, but the derivations are not very important. Students must remember all the results as the questions can't be solved unless they know these results<\/p>\n

             <\/p>\n

            Example:<\/strong> Evaluate \"\"<\/p>\n

            Solution:<\/strong> x3<\/sup> – 1 ≤ 0 on [–1, 0]<\/p>\n

            x3<\/sup> – 1 ≤ 0 on [0, 1]<\/p>\n

            x3<\/sup> – 1 ≥ 0 on [1, 2]<\/p>\n

            \"\"  <\/p>\n

                              = \"\"<\/p>\n

                              = \"\"<\/p>\n

                              = \"\"<\/p>\n

                              = \"\"<\/p>\n

             <\/p>\n

            Antiderivatives at Specific Integral<\/u><\/strong>: –<\/p>\n

             <\/p>\n

            Antiderivatives, also known as primitives, are functions that are obtained by reversing the process of differentiation. They allow us to find the original function given its derivative.<\/p>\n

             <\/p>\n

            When evaluating antiderivatives at a specific interval, it usually involves finding the difference between the antiderivative values at the upper and lower limits of the interval. This process is commonly used in definite integration to calculate the net area under a curve.<\/p>\n

             <\/p>\n

            Here's an example to illustrate the concept:<\/p>\n

             <\/p>\n

            Let's consider the function f(x) = 2x, and we want to find the antiderivative (or primitive) of this function. Integrating the function for x, we obtain: –<\/p>\n

            ∫ 2x dx = x2<\/sup> + C<\/p>\n

             <\/p>\n

            Here, C represents the constant of integration, which arises because differentiation "forgets" about the constant term in a function. When dealing with definite integrals, this constant of integration cancels out when subtracting the values at the upper and lower limits, so it doesn't affect the final result.<\/p>\n

             <\/p>\n

            Now, if we want to find the value of the antiderivative within a specific interval, say, between x = 1 and x = 3, we can evaluate the antiderivative at these limits:<\/p>\n

             <\/p>\n

            F(3) – F(1) = (32<\/sup> + C) – (12<\/sup> + C) = 9 + C – 1 – C = 8<\/p>\n

             <\/p>\n

            So, the value of the antiderivative at the interval [1, 3] is 8.<\/p>\n

             <\/p>\n

            In this example, we observe that the constant of integration cancels out when subtracting the antiderivative values at the upper and lower limits, resulting in a definite integral value. However, it's important to note that the presence of a constant of integration may affect the value of the antiderivative at a specific point, but it doesn't affect the net area under the curve within a given interval.          <\/p>\n

            Key Points<\/u><\/strong><\/p>\n

              \n
            • The definite integral represents the net area under the curve of a function within a specific interval.<\/li>\n<\/ul>\n

               <\/p>\n

                \n
              • By evaluating a definite integral, we can find the value of a function over a given interval.<\/li>\n<\/ul>\n

                 <\/p>\n

                  \n
                • The definite integral is denoted by ∫[a, b] f(x) dx, where 'a' and 'b' represent the lower and upper limits of the interval, respectively, and f(x) is the function to be integrated.<\/li>\n<\/ul>\n

                   <\/p>\n

                    \n
                  • To find the value of a function using a definite integral, we need to calculate the antiderivative (or primitive) of the function.<\/li>\n<\/ul>\n

                     <\/p>\n

                      \n
                    • The antiderivative of a function is obtained by reversing the process of differentiation.<\/li>\n<\/ul>\n

                       <\/p>\n

                        \n
                      • The constant of integration, denoted by 'C', is included in the antiderivative, but when evaluating a definite integral, it cancels out when subtracting the values at the upper and lower limits.<\/li>\n<\/ul>\n

                         <\/p>\n

                          \n
                        • Evaluating a definite integral involves substituting the upper limit into the antiderivative and subtracting the result at the lower limit.<\/li>\n<\/ul>\n

                           <\/p>\n

                            \n
                          • The result of the definite integral represents the net area under the curve of the function within the specified interval and can be interpreted as the value of the function.<\/li>\n<\/ul>\n

                             <\/p>\n

                              \n
                            • Definite integrals are widely used in various fields, such as physics, engineering, and economics, to calculate quantities such as displacement, velocity, work, and total profit.<\/li>\n<\/ul>\n

                               <\/p>\n

                                \n
                              • Numerical methods, such as the trapezoidal rule or Simpson's rule, can be used when it is not possible to find an exact antiderivative or when dealing with complex functions, to approximate the value of a definite integral.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

                                Unit: Application of Integrations Chapter: Value of Function by Definite Integral Reference: – Differentiability & Continuity, Increasing & Decreasing functions, Curve sketching, Analysis of a function, Optimization problems, First & Second Derivative test, Related rates, Local extrema, Implicit differentiation & Applications.   After studying this chapter, you should be able to: Identifying the Function & […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[627],"tags":[],"class_list":["post-9350","post","type-post","status-publish","format-standard","hentry","category-ap-calculus-bc"],"_links":{"self":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9350","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/comments?post=9350"}],"version-history":[{"count":0,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9350\/revisions"}],"wp:attachment":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/media?parent=9350"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/categories?post=9350"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/tags?post=9350"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}