What is the integral from-3 to 1 of [(2x-2)(2x+2)] ?

The integral from-3 to 1 of [(2x-2)(2x+2)] is 64/3

Popular >

integral from-3 to 1 of [(2x-2)(2x+2)]

\int_{- 3}^{1} [(2 x - 2) (2 x + 2)] d x

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

\int_{- 3}^{1} (2 x - 2) (2 x + 2) d x

Apply Integration By Parts

: [(2 x + 2) (x^{2} - 2 x) - \int 2 (x^{2} - 2 x) d x]_{- 3}^{1}

= [(2 x + 2) (x^{2} - 2 x) - \int 2 (x^{2} - 2 x) d x]_{- 3}^{1}

\int 2 (x^{2} - 2 x) d x = 2 (\frac{x ^{3}}{3} - x^{2})

= [(2 x + 2) (x^{2} - 2 x) - 2 (\frac{x ^{3}}{3} - x^{2})]_{- 3}^{1}

Simplify

[(2 x + 2) (x^{2} - 2 x) - 2 (\frac{x ^{3}}{3} - x^{2})]_{- 3}^{1} : [\frac{1}{3} (4 x^{3} - 12 x)]_{- 3}^{1}

= [\frac{1}{3} (4 x^{3} - 12 x)]_{- 3}^{1}

Compute the boundaries:

\frac{64}{3}

= \frac{64}{3}

\int_{1}^{x^{5}} lo g_{3} (t + 2) d t

\int_{2}^{0} sin (x) d x

\int_{4}^{5} \frac{2}{x ^{2} - 1} d x

\int_{- 1}^{0} (x^{2} - 2 x - \frac{1}{3}) d x

\int_{\frac{1}{2}}^{\frac{3}{2}} (x + \frac{1}{x}) d x

What is the integral from-3 to 1 of [(2x-2)(2x+2)] ?
The integral from-3 to 1 of [(2x-2)(2x+2)] is 64/3