What is the integral from 1 to 2 of 15x^2(x^3-7)^4 ?

The integral from 1 to 2 of 15x^2(x^3-7)^4 is 7777

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of 15x^2(x^3-7)^4

\int_{1}^{2} 15 x^{2} (x^{3} - 7)^{4} d x

7777

Solution steps

\int_{1}^{2} 15 x^{2} (x^{3} - 7)^{4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 15 \cdot \int_{1}^{2} x^{2} (x^{3} - 7)^{4} d x

Apply u-substitution

: \int_{- 6}^{1} \frac{u ^{4}}{3} d u

= 15 \cdot \int_{- 6}^{1} \frac{u ^{4}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 15 \cdot \frac{1}{3} \cdot \int_{- 6}^{1} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 6}^{1}

= 15 \cdot \frac{1}{3} [\frac{u ^{5}}{5}]_{- 6}^{1}

Simplify

15 \cdot \frac{1}{3} [\frac{u ^{5}}{5}]_{- 6}^{1} : 5 [\frac{u ^{5}}{5}]_{- 6}^{1}

= 5 [\frac{u ^{5}}{5}]_{- 6}^{1}

Compute the boundaries:

\frac{7777}{5}

= 5 \cdot \frac{7777}{5}

Simplify

= 7777

\int_{0}^{10} 4 x d x

\int_{0}^{4} 2 e^{x} d x

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

\int_{0}^{1} 6 t 4 t^{2} + 1 d t

\int_{1}^{6} x^{2} - 7 x + 6 d x

What is the integral from 1 to 2 of 15x^2(x^3-7)^4 ?
The integral from 1 to 2 of 15x^2(x^3-7)^4 is 7777