What is the integral from 0 to 2 of (2x-3)(8x^2+2) ?

The integral from 0 to 2 of (2x-3)(8x^2+2) is -4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (2x-3)(8x^2+2)

\int_{0}^{2} (2 x - 3) (8 x^{2} + 2) d x

- 4

Solution steps

\int_{0}^{2} (2 x - 3) (8 x^{2} + 2) d x

Apply Integration By Parts

: [\frac{16 x ^{4}}{3} + 4 x^{2} - 8 x^{3} - 6 x - \int 2 (\frac{8 x ^{3}}{3} + 2 x) d x]_{0}^{2}

= [\frac{16 x ^{4}}{3} + 4 x^{2} - 8 x^{3} - 6 x - \int 2 (\frac{8 x ^{3}}{3} + 2 x) d x]_{0}^{2}

\int 2 (\frac{8 x ^{3}}{3} + 2 x) d x = 2 (\frac{2 x ^{4}}{3} + x^{2})

= [\frac{16 x ^{4}}{3} + 4 x^{2} - 8 x^{3} - 6 x - 2 (\frac{2 x ^{4}}{3} + x^{2})]_{0}^{2}

Simplify

[\frac{16 x ^{4}}{3} + 4 x^{2} - 8 x^{3} - 6 x - 2 (\frac{2 x ^{4}}{3} + x^{2})]_{0}^{2} : [2 x (2 x^{3} - 4 x^{2} + x - 3)]_{0}^{2}

= [2 x (2 x^{3} - 4 x^{2} + x - 3)]_{0}^{2}

Compute the boundaries:

- 4

= - 4

\int_{\frac{π}{4}}^{\frac{5 π}{4}} cos (x) d x

\int_{- 2}^{5} \frac{2}{3} d x

\int_{0}^{2 π} \frac{128}{15} d θ

\int_{3}^{7} 2 x d x

\int_{1}^{1.5} (2 x + \frac{3}{2 x ^{2}}) d x

What is the integral from 0 to 2 of (2x-3)(8x^2+2) ?
The integral from 0 to 2 of (2x-3)(8x^2+2) is -4