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

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

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integral from 0 to 2 of (x^3+1)^{1/2}x^2

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

\frac{52}{9}

Decimal Notation

5.77777 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{8} \frac{( u + 1 ) ^{\frac{1}{2}}}{3} d u

= \int_{0}^{8} \frac{( u + 1 ) ^{\frac{1}{2}}}{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_{0}^{8} (u + 1)^{\frac{1}{2}} d u

Apply u-substitution

: \int_{1}^{9} v^{\frac{1}{2}} d v

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{52}{3}

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

Simplify

= \frac{52}{9}

\int 3 \frac{sin ( 2 x )}{sin ( x )} d x

(- 5, - 2), (5, 14)

x \to - \frac{1}{4} lim (2)

x \to a lim ([2 f (3) - 3 g (4)])

y^{'} (t)

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