# How to find a bound integral?

## Question:

How to find a bound integral?

## Definite Integral or Bounded Integral :

Suppose we have any function {eq}f(x) {/eq} of one variable. Then using the concept of definite integral we can write:

{eq}\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right) {/eq}

Where, {eq}F {/eq} is the primitive or antiderivative of {eq}f {/eq}

Again we know that: {eq}\displaystyle \int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \ \ \ \ \ \ \ \text{(where C is the constant of integration)} {/eq}

## Answer and Explanation:

Let us give an example of bound integral as below manner:

{eq}\eqalign{ I& = \int\limits_2^3 {{{\left( {1 + \frac{1}{x}} \right)}^2}} dx \cr & = \int\limits_2^3 {\left( {1 + \frac{2}{x} + \frac{1}{{{x^2}}}} \right)} dx\,\,\,\,\,\,\left[ {{\text{As we know: }}{{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}} \right] \cr & = \int\limits_2^3 {\left( {1 + 2{x^{ - 1}} + {x^{ - 2}}} \right)} dx \cr & = \int\limits_2^3 {dx} + \int\limits_2^3 {2{x^{ - 1}}dx} + \int\limits_2^3 {{x^{ - 2}}dx} \cr & = \left[ x \right]_2^3 + \left[ {2\ln \left( x \right)} \right]_2^3 + \left[ {\frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_2^3 \cr & = \left[ {3 - 2} \right] + \left[ {2\ln \left( 3 \right) - 2\ln \left( 2 \right)} \right] + \left[ {\frac{{{3^{ - 1}}}}{{ - 1}} - \frac{{{2^{ - 1}}}}{{ - 1}}} \right] \cr & = \color{blue}{\ln \left( {\frac{9}{4}} \right) + \frac{7}{6}} \cr} {/eq}

Hence we can easily say that using the bound integral, one can find the area of the region under the curve.

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from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2