Question

Evaluate the integral

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

Short answer

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

Step-by-step solution

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 \({{\rm{F}}^{\rm{'}}}{\rm{ = f}}\)

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.

Using Evaluation theorem

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