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

The integral from 0 to 2 of 2x^2sqrt(x^3+1) is 104/9

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

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

\frac{104}{9}

Decimal Notation

11.55555 \dots

Solution steps

\int_{0}^{2} 2 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

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

Apply u-substitution

: \int_{1}^{9} \frac{u}{3} d u

= 2 \cdot \int_{1}^{9} \frac{u}{3} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int_{1}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= 2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Simplify

2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9} : \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{2}{3} \cdot \frac{52}{3}

Simplify

= \frac{104}{9}

y^{^{''}} + 16 y = 24 sec (4 t)

x \to - 4 + lim (f (x))

x^{2} + x y + 2 y^{2} = 32, (5, 1)

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

What is the integral from 0 to 2 of 2x^2sqrt(x^3+1) ?
The integral from 0 to 2 of 2x^2sqrt(x^3+1) is 104/9