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

The integral from-1 to 1 of 3x^2sqrt(x^3+1) is (4sqrt(2))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of 3x^2sqrt(x^3+1)

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

\frac{4 2}{3}

Decimal Notation

1.88561 \dots

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{- 1}^{1} x^{2} x^{3} + 1 d x

Apply u-substitution

: \int_{0}^{2} \frac{u}{3} d u

= 3 \cdot \int_{0}^{2} \frac{u}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{0}^{2} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{2}

= 3 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{2}

Simplify

3 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{2} : [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{2}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{2}

Compute the boundaries:

\frac{4 2}{3}

= \frac{4 2}{3}

\frac{d}{d x} ((1 + x^{4})^{\frac{2}{3}})

(4) (8.6)

f (x) = (4 x + 1)

\frac{\partial}{\partial x} (ln (3 x + 5 y))

\frac{d}{d x} ((2 x)^{2})

What is the integral from-1 to 1 of 3x^2sqrt(x^3+1) ?
The integral from-1 to 1 of 3x^2sqrt(x^3+1) is (4sqrt(2))/3