What is the integral from 1 to 3 of (4x-3)^2-(x^2)^2 ?

The integral from 1 to 3 of (4x-3)^2-(x^2)^2 is 184/15

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integral from 1 to 3 of (4x-3)^2-(x^2)^2

\int_{1}^{3} (4 x - 3)^{2} - (x^{2})^{2} d x

\frac{184}{15}

Decimal Notation

12.26666 \dots

Solution steps

\int_{1}^{3} (4 x - 3)^{2} - (x^{2})^{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_{1}^{3} (4 x - 3)^{2} d x - \int_{1}^{3} (x^{2})^{2} d x

\int_{1}^{3} (4 x - 3)^{2} d x = \frac{182}{3}

\int_{1}^{3} (x^{2})^{2} d x = \frac{242}{5}

= \frac{182}{3} - \frac{242}{5}

Simplify

\frac{182}{3} - \frac{242}{5} : \frac{184}{15}

= \frac{184}{15}

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What is the integral from 1 to 3 of (4x-3)^2-(x^2)^2 ?
The integral from 1 to 3 of (4x-3)^2-(x^2)^2 is 184/15