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

The integral from 1 to 2 of (x^2-1)^2 is 38/15

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integral from 1 to 2 of (x^2-1)^2

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

\frac{38}{15}

Decimal

2.53333 \dots

Solution steps

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

Expand

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

= \int_{1}^{2} x^{4} - 2 x^{2} + 1 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_{1}^{2} x^{4} d x - \int_{1}^{2} 2 x^{2} d x + \int_{1}^{2} 1 d x

\int_{1}^{2} x^{4} d x = \frac{31}{5}

\int_{1}^{2} 2 x^{2} d x = \frac{14}{3}

\int_{1}^{2} 1 d x = 1

= \frac{31}{5} - \frac{14}{3} + 1

Simplify

\frac{31}{5} - \frac{14}{3} + 1 : \frac{38}{15}

= \frac{38}{15}

What is the integral from 1 to 2 of (x^2-1)^2 ?
The integral from 1 to 2 of (x^2-1)^2 is 38/15