Question

Evaluate the integral.

\(\int\limits_{\rm{0}}^{\rm{3}} {{\rm{(1 + 6}}{{\rm{w}}^{\rm{2}}}{\rm{ - 10}}{{\rm{w}}^{\rm{4}}}{\rm{)dw}}} \)

Short answer

The value of the integral is \({\rm{ - 429}}\).

Step-by-step solution

Defining Evaluation Theorem

To find the general antiderivative, using the indefinite integral

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

\(\begin{aligned}{c}\int {\left( {{\rm{1 + 6}}{{\rm{w}}^{\rm{2}}}{\rm{ - 10}}{{\rm{w}}^{\rm{4}}}} \right){\rm{dw = 6 \times }}\frac{{{{\rm{w}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - 10}}} {\rm{ \times }}\frac{{{{\rm{w}}^{\rm{5}}}}}{{\rm{5}}}{\rm{ + C}}\\\end{aligned}\)

\({\rm{ = w + 2w}}{{\rm{t}}^{\rm{3}}}{\rm{ - 2}}{{\rm{w}}^{\rm{5}}}{\rm{ + C = F(w)}}\)

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

Assigning the values

Therefore from the evaluation theorem,

\(\begin{aligned}{c}\int {\left( {{\rm{1 + 6}}{{\rm{w}}^{\rm{2}}}{\rm{ - 10}}{{\rm{w}}^{\rm{4}}}} \right){\rm{dw = }}\left( {{\rm{w + 2}}{{\rm{w}}^{\rm{3}}}{\rm{ - 2}}{{\rm{w}}^{\rm{5}}}} \right)_{\rm{0}}^{\rm{3}}} \\\end{aligned}\)

\({\rm{ = }}\left( {{\rm{3 + 2(3}}{{\rm{)}}^{\rm{3}}}{\rm{ - 2(3}}{{\rm{)}}^{\rm{5}}}} \right){\rm{ - }}\left( {{\rm{0 + 0 - 0}}} \right)\)

\({\rm{ = }}\left( {{\rm{ - 429 - 0}}} \right){\rm{ = }}\left( {{\rm{ - 429}}} \right)\)