What is the integral from-2 to 3 of (4x-3)(x-3) ?

The integral from-2 to 3 of (4x-3)(x-3) is 325/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 3 of (4x-3)(x-3)

\int_{- 2}^{3} (4 x - 3) (x - 3) d x

\frac{325}{6}

Solution steps

\int_{- 2}^{3} (4 x - 3) (x - 3) d x

Apply Integration By Parts

: [(4 x - 3) (\frac{x ^{2}}{2} - 3 x) - \int 4 (\frac{x ^{2}}{2} - 3 x) d x]_{- 2}^{3}

= [(4 x - 3) (\frac{x ^{2}}{2} - 3 x) - \int 4 (\frac{x ^{2}}{2} - 3 x) d x]_{- 2}^{3}

\int 4 (\frac{x ^{2}}{2} - 3 x) d x = 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{2})

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

Simplify

[(4 x - 3) (\frac{x ^{2}}{2} - 3 x) - 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{2})]_{- 2}^{3} : [\frac{1}{6} (8 x^{3} - 45 x^{2} + 54 x)]_{- 2}^{3}

= [\frac{1}{6} (8 x^{3} - 45 x^{2} + 54 x)]_{- 2}^{3}

Compute the boundaries:

\frac{325}{6}

= \frac{325}{6}

\int_{- 1}^{1} (x^{5} - x^{6}) d x

\int_{1}^{3} k x^{2} - 2 d x

\int_{0}^{2} - 2 x + 2 d x

\int_{0}^{1} \frac{1}{1 + 2 x} d x

\int_{0}^{6 - x - z} 1 d y

What is the integral from-2 to 3 of (4x-3)(x-3) ?
The integral from-2 to 3 of (4x-3)(x-3) is 325/6