What is the integral from 1 to 2 of (x^2-3)^2 ?

The integral from 1 to 2 of (x^2-3)^2 is 6/5

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integral from 1 to 2 of (x^2-3)^2

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

\frac{6}{5}

Decimal Notation

1.2

Solution steps

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

Expand

(x^{2} - 3)^{2} : x^{4} - 6 x^{2} + 9

= \int_{1}^{2} x^{4} - 6 x^{2} + 9 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} 6 x^{2} d x + \int_{1}^{2} 9 d x

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

\int_{1}^{2} 6 x^{2} d x = 14

\int_{1}^{2} 9 d x = 9

= \frac{31}{5} - 14 + 9

Simplify

\frac{31}{5} - 14 + 9 : \frac{6}{5}

= \frac{6}{5}

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What is the integral from 1 to 2 of (x^2-3)^2 ?
The integral from 1 to 2 of (x^2-3)^2 is 6/5