What is the integral from-1 to 0 of 3x^2sqrt(x^3+1) ?

The integral from-1 to 0 of 3x^2sqrt(x^3+1) is 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 0 of 3x^2sqrt(x^3+1)

\int_{- 1}^{0} 3 x^{2} x^{3} + 1 d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{- 1}^{0} 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}^{0} x^{2} x^{3} + 1 d x

Apply u-substitution

: \int_{0}^{1} \frac{u}{3} d u

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

\int_{0}^{2} π x^{2} d x

\int_{2}^{\infty} e^{- s t} d t

\int_{\frac{π}{6}}^{\frac{π}{2}} csc (x) d x

\int_{7}^{8} (\frac{5}{x ^{2}} - 1) d x

\int_{4}^{7} x d x

What is the integral from-1 to 0 of 3x^2sqrt(x^3+1) ?
The integral from-1 to 0 of 3x^2sqrt(x^3+1) is 2/3