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### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integral$$\int\limits_{{\rm{ - 2}}}^{\rm{0}} {\left( {{\rm{(}}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{t}}^{\rm{4}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{4}}}{{\rm{t}}^{\rm{3}}}{\rm{ - t}}} \right)} {\rm{dt}}$$ 

The value of the integral is $$\frac{{{\rm{21}}}}{{\rm{5}}}$$

See the step by step solution

## Step 1: Evaluation Theorem

If $${\rm{f}}$$is continuous on the interval $$\left( {{\rm{a,b}}} \right)$$, then $$\int\limits_{\rm{a}}^{\rm{b}} {{\rm{f(x)dx = F(b) - F(a)}}}$$

Where $${\rm{F}}$$ is any antiderivative of $${\rm{f}}$$, that is $${{\rm{F}}^{\rm{'}}}{\rm{ = f}}$$

## Step 2: Cancels out arbitrary constant C

To find the general antiderivative, using the indefinite integral



and ignoring (just for the moment) the boundaries of integration.

We will leave out the arbitrary constant C, since it cancels out when calculating a definite integral.

## Step 3: Using Evaluation theorem

$$\begin{array}{c}\int {{\rm{(}}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{t}}^{\rm{4}}}} {\rm{ + }}\frac{{\rm{1}}}{{\rm{4}}}{{\rm{t}}^{\rm{3}}}{\rm{ - t)dt = }}\left( {\frac{{\rm{1}}}{{{\rm{10}}}}{{\rm{t}}^{\rm{5}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{\rm{t}}^{\rm{4}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{t}}^{\rm{2}}}} \right]_{{\rm{ - 2}}}^{\rm{0}}\\{\rm{ &= }}\left( {\frac{{\rm{1}}}{{{\rm{10}}}}{{{\rm{(0)}}}^{\rm{5}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{{\rm{(0)}}}^{\rm{4}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(0)}}} \right){\rm{ - }}\left( {\frac{{\rm{1}}}{{{\rm{10}}}}{{{\rm{( - 2)}}}^{\rm{5}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{{\rm{( - 2)}}}^{\rm{4}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}{{{\rm{( - 2)}}}^{\rm{2}}}} \right)\\ &= - \left( {\frac{{32}}{{10}} + 1 - 2} \right) = - \left( { - \frac{{42}}{{10}}} \right) = \frac{{21}}{5}\end{array}$$