Question

Evaluate the integral

\(\int\limits_{\rm{0}}^{\rm{2}} {{\rm{(2x - 3)(4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 1)dx}}} \)

Short answer

The value of the integral is\({\rm{ = - 2}}\)

Step-by-step solution

Expanding the integrand.

By expanding the integrand

\(\begin{array}{c}{\rm{(2x - 3)(4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 1) = 2x(4}}{{\rm{x}}^{\rm{2}}}{\rm{) - 2x(1) - 3(4}}{{\rm{x}}^{\rm{2}}}{\rm{) - 3(1)}}\\{\rm{ = 8}}{{\rm{x}}^{\rm{2}}}{\rm{ + 2x - 12}}{{\rm{x}}^{\rm{2}}}{\rm{ - 3}}\end{array}\)

Therefore we have to evaluate

\(\int\limits_{\rm{0}}^{\rm{2}} {\rm{(}} {\rm{8}}{{\rm{x}}^{\rm{2}}}{\rm{ + 2x - 12}}{{\rm{x}}^{\rm{2}}}{\rm{ - 3)dx}}\)

General antiderivative

Find the general antiderivative using the indefinite integral

And ignore (just for the moment) the boundaries of integration.

\(\begin{aligned}{c}\int {{\rm{(8}}{{\rm{x}}^{\rm{3}}}{\rm{ + 2x - 12}}{{\rm{x}}^{\rm{2}}}{\rm{ - 3)dx}}} {\rm{ = 8 \times }}\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}{\rm{ + 2 \times }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - 12 \times }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - 3x + C}}\\{\rm{ = 2}}{{\rm{x}}^{\rm{4}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}{\rm{ - 4}}{{\rm{x}}^{\rm{3}}}{\rm{ - 3x + C = F(x)}}\end{aligned}\)

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

Assigning final value

\(\begin{aligned}{c}\int\limits_{\rm{0}}^{\rm{2}} {\rm{(}} {\rm{8}}{{\rm{x}}^{\rm{2}}}{\rm{ + 2x - 12}}{{\rm{x}}^{\rm{2}}}{\rm{ - 3)dx = }}\left( {{\rm{2}}{{\rm{x}}^{\rm{4}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}{\rm{ - 4}}{{\rm{x}}^{\rm{3}}}{\rm{ - 3x}}} \right)_{\rm{0}}^{\rm{2}}{\rm{ = }}\left( {{\rm{2 \times }}{{\rm{2}}^{\rm{4}}}{\rm{ \times }}{{\rm{2}}^{\rm{2}}}{\rm{ - 4 \times }}{{\rm{2}}^{\rm{3}}}{\rm{ - 3 \times 2}}} \right){\rm{ - 0}}\\{\rm{ = - 2 - 0 = - 2}}\end{aligned}\)