# Evaluate the integrals: (a) integral_7^8 (8 / {x^2} - 1) dx. (b) integral_0^4 (3 - t) square root...

## Question:

Evaluate the integrals:

(a) {eq}\displaystyle \int_7^8 \bigg(\dfrac 8 {x^2} - 1\bigg)\ dx {/eq}.

(b) {eq}\displaystyle \int_0^4 (3 - t) \sqrt t\ dt {/eq}.

(c) {eq}\displaystyle \int_2^7 (3 - |x - 4|)\ dx {/eq}.

## Solving Integrals by Power Rule:

When dealing with integrals of polynomial functions, the power rule method is a straightforward technique that can be used. This simple rule is given by the following formula

{eq}\displaystyle \int ax^n \, dx = \frac{ax^{n+1}}{n+1} + C {/eq}.

## Answer and Explanation:

(a)

{eq}\displaystyle \begin{align} \int_7^8 \bigg(\dfrac 8 {x^2} - 1\bigg) \, dx & = \int_7^8 \bigg( 8x^{-2} - 1\bigg) \, dx \\ & = \bigg[ \frac{8x^{-2+1}}{-2+1} - x\bigg]_7^8\\ & = \bigg[ \frac{8x^{-1}}{-1} - x\bigg]_7^8\\ & = \bigg[ - \frac{8}{x} - x\bigg]_7^8\\ & = - \frac{8}{8} - 8 + frac{8}{7} + 7 \\ & = - 1 - 8 + \frac{8}{7} + 7 \\ & = - \frac{6}{7} \\ \end{align} {/eq}

(b)

{eq}\displaystyle \begin{align} \int_0^4 (3 - t) \sqrt t \, dt &= \int_0^4 (3 - t) t^{\frac{1}{2}} \, dt \\ &= \int_0^4 ( 3 t^{\frac{1}{2}} - t^{\frac{3}{2}} )\, dt \\ &= \bigg[ \frac{ 3 t^{\frac{1}{2} + 1} }{\frac{1}{2} + 1} - \frac{ t^{\frac{3}{2}+1} }{\frac{3}{2}+1} \bigg]_0^4 \\ &= \bigg[ \frac{ 3 t^{\frac{3}{2} } }{\frac{3}{2}} - \frac{ t^{\frac{5}{2}} }{\frac{5}{2}} \bigg]_0^4 \\ &= \bigg[ 2 t^{\frac{3}{2} } - \frac{2}{5} t^{\frac{5}{2}} \bigg]_0^4 \\ &= 2 \cdot 4^{\frac{3}{2} } - \frac{2}{5} \cdot 4^{\frac{5}{2}} \\ &= 2 \cdot 8 - \frac{2}{5} \cdot 32 \\ &= 16 - \frac{64}{5} \\ &= \frac{16}{5} \\ \end{align} {/eq}

(c)

{eq}\displaystyle \begin{align} \int_2^7 (3 - |x - 4|) \, dx &= \int_2^4 [ 3 - (-x + 4) ] \, dx + \int_4^7 [ (3 - (x - 4) ] \, dx &\text{ Truncate the integral to eliminate the absolutes. } \\ &= \int_2^4 ( 3 + x - 4) \, dx + \int_4^7 (3 - x + 4) \, dx \\ &= \int_2^4 ( x - 1) \, dx + \int_4^7 (7 - x ) \, dx \\ &= \bigg[ \frac{x^2}{2} - x \bigg]_2^4 + \bigg[ 7x - \frac{x^2}{2} \bigg]_4^7 \\ &= \bigg[ \frac{1}{2}(4^2 -2^2) - (4-2) \bigg]_ + \bigg[ 7(7-4) - \frac{1}{2 } (7^2 -4^2) \bigg] \\ &= \bigg[ \frac{1}{2}(12) - 2 \bigg] + \bigg[ 7(3) - \frac{1}{2 } (33) \bigg] \\ &= 4 + 21 - \frac{33}{2 }\\ & = \frac{17}{2} \end{align} {/eq}

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13