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

The integral from 0 to 2 of x^2(x^2+1) is 136/15

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

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

\frac{136}{15}

Decimal

9.06666 \dots

Solution steps

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

Expand

x^{2} (x^{2} + 1) : x^{4} + x^{2}

= \int_{0}^{2} x^{4} + x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2} x^{4} d x + \int_{0}^{2} x^{2} d x

\int_{0}^{2} x^{4} d x = \frac{32}{5}

\int_{0}^{2} x^{2} d x = \frac{8}{3}

= \frac{32}{5} + \frac{8}{3}

Simplify

\frac{32}{5} + \frac{8}{3} : \frac{136}{15}

= \frac{136}{15}

\int_{3}^{5} (2 x - 5)^{4} d x

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

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

\int_{0}^{3} \frac{1}{3 x} d x

\int_{0}^{4} \frac{1}{2} (4 - x)^{2} d x

What is the integral from 0 to 2 of x^2(x^2+1) ?
The integral from 0 to 2 of x^2(x^2+1) is 136/15