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Definite Integrals with Limits as Functions

Go back to  'Definite Integration'

(14)    In the unit on “Integration Basics”, we saw that for a function \(f(x)\) , the anti-derivative \(g(x)\) was defined as

\[\begin{align}&g(x) = \int\limits_a^x {f(t)dt} \qquad \quad ({\rm{where }}\;a\;{\rm{ is\; a\; constant)}}\\\text{so that} \qquad \quad &g'(x) = f(x)\end{align}\]

We now consider an integral of the following form:

\[h(x) = \int\limits_{\phi (x)}^{\psi (x)} {f(t)dt} \]

That is, the limits of integration are themselves functions of x. The anti-derivative \(g(x)\) is a special case of \(h(x)\) with \(\psi (x) = x\) and \(\phi (x) = a.\)

Now, how do we evaluate \(h'(x)?\) Leibnitz’s rule for differentiation tells us how to do so. Since \(g(x)\) is the anti-derivative of \(f(x),\) we have:

\[\begin{align}&\;\;\;\;\,\,\,\,\,\,\,\,h(x) = \int\limits_{\phi (x)}^{\psi (x)} {f(t)dt} \\&\qquad\quad\;\;\;\;= \left. {g(t)} \right|_{\phi (x)}^{\psi (x)}\\\\&\qquad\quad\;\;\;\; = g\left( {\psi (x)} \right) - g\left( {\phi (x)} \right)\\\\&\Rightarrow \,\,\,\,h'(x) = g'\left( {\psi (x)} \right)\psi '(x) - g'\left( {\phi (x)} \right)\phi '(x)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\;\;\;\;\;= f\left( {\psi (x)} \right)\psi '(x) - f\left( {\phi (x)} \right)\phi '(x).\end{align}\]

Let us see an example of this rule:

Example –20

Evaluate \(f'(x)\) if \(f(x) = \int\limits_x^{{x^2}} {\left( {{t^2} + 1} \right)dt} \)

Solution: Let us first find out \(f(x)\) using straight forward integration:

\[\begin{align} f(x) &= \left. {\left( {\frac{{{t^3}}}{3} + t} \right)} \right|_x^{{x^2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{{x^6}}}{3} + {x^2} - \frac{{{x^3}}}{3} - x\\\Rightarrow \quad f'(x) &= 2{x^5} - {x^2} + 2x - 1\end{align}\]

Now we redo this using the Leibnitz’s differentiation rule:

\[\begin{array}{l}f'(x) = \left( {{{\left( {{x^2}} \right)}^2} + 1} \right){\left( {{x^2}} \right)^\prime } - \left( {{{\left( x \right)}^2} + 1} \right){\left( x \right)^\prime }\\\,\,\,\,\,\,\,\,\,\,\,\,\, = ({x^4} + 1)(2x) - ({x^2} + 1)(1)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{x^5} - {x^2} + 2x - 1\end{array}\]

Let us now do an example of this rule where straight forward integration would be much more difficult.

Example –21

Determine the equation of the tangent to the curve \(y = f(x)\) at x = 1, where

\[\begin{align}f(x) = \int\limits_{{x^2}}^{{x^3}} {\frac{1}{{\sqrt {1 + {t^5}} }}dt}\end{align}\]

Solution: Notice how tedious it would be to actually carry out the integration. Instead, we use the Leibnitz’s differentiation rule:

\[\begin{align}&\;\;\;\,\,\,\,\,\,f'(x) = \frac{{3{x^2}}}{{\sqrt {1 + {x^{15}}} }} - \frac{{2x}}{{\sqrt {1 + {x^{10}}} }}\\ &\Rightarrow \,f'(1) = \frac{3}{{\sqrt 2 }} - \frac{2}{{\sqrt 2 }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad= \frac{1}{{\sqrt 2 }}\end{align}\]

Also, at x = 1,

\[\begin{align}&f(x) = f(1) = \int\limits_1^1 {\frac{1}{{\sqrt {1 + {t^5}} }}dt} \\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad= 0\end{align}\]

Thus, the tangent passes through (1, 0) and has slope \(\frac{1}{{\sqrt 2 }}\) . The required equation is

\[\begin{align}&\qquad\;\;\,\,\,\,\,\,\,\,y - 0 = \frac{1}{{\sqrt 2 }}(x - 1)\\ &\Rightarrow \qquad\,\,x - \sqrt 2 y - 1 = 0\end{align}\]

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Definite Integration
grade 11 | Answers Set 2
Definite Integration
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Definite Integration
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Definite Integration
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Download SOLVED Practice Questions of Definite Integrals with Limits as Functions for FREE
Definite Integration
grade 11 | Answers Set 2
Definite Integration
grade 11 | Questions Set 1
Definite Integration
grade 11 | Answers Set 1
Definite Integration
grade 11 | Questions Set 2
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