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

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

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

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

23 - \frac{2}{3}

Decimal Notation

2.79743 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

23 - \frac{2}{3}

= 23 - \frac{2}{3}

\frac{\partial}{\partial x} (4 x y - x^{4} - y^{4})

y = x^{3}, y = 2 x - x^{2}

x \to 2 lim (\frac{( x ^{2} + 7 x - 44 )}{x ^{2} - 6 x + 8})

\int_{- \infty}^{- 1} e^{- 34 x} d x

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