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

The integral from 0 to 2 of x^2sqrt(x^3+1) is 52/9

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

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

\frac{52}{9}

Decimal Notation

5.77777 \dots

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{52}{3}

= \frac{1}{3} \cdot \frac{52}{3}

Simplify

= \frac{52}{9}

f (x) = \frac{7}{4 - x}, at x = - 1

\frac{\partial}{\partial x} (y^{4} sin (2 x))

x \to - 4 lim (\frac{x ^{2} - 9}{4 - x})

f (x) = 6 e^{- \frac{1}{6} x^{2}}

\int 24 y (1 - x - y) d y

What is the integral from 0 to 2 of x^2sqrt(x^3+1) ?
The integral from 0 to 2 of x^2sqrt(x^3+1) is 52/9