What is the integral from-1 to 1 of 3x^2sqrt(x^3+5) ?

The integral from-1 to 1 of 3x^2sqrt(x^3+5) is 4sqrt(6)-16/3

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integral from-1 to 1 of 3x^2sqrt(x^3+5)

\int_{- 1}^{1} 3 x^{2} x^{3} + 5 d x

46 - \frac{16}{3}

Solution steps

\int_{- 1}^{1} 3 x^{2} x^{3} + 5 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} + 5 d x

Apply u-substitution

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

= 3 \cdot \int_{4}^{6} \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_{4}^{6} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{6}

= 3 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{6}

Simplify

3 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{6} : [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{6}

= [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{6}

Compute the boundaries:

46 - \frac{16}{3}

= 46 - \frac{16}{3}

(x^{2} + 2 x)^{^{'}}

x \to 1 + lim (\frac{5}{x ^{3} - 1})

\int \frac{x + 5}{x ^{2} + 10 x} d x

\int_{0}^{1} π (1 - x^{2}) d x

\frac{- 5}{s ( s ^{2} - 5 s + 4 )}

What is the integral from-1 to 1 of 3x^2sqrt(x^3+5) ?
The integral from-1 to 1 of 3x^2sqrt(x^3+5) is 4sqrt(6)-16/3