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Essential Calculus: Early Transcendentals
Found in: Page 289
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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 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

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

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