What is the integral from 0 to 1 of (3x)^2-(3x^2)^2 ?

The integral from 0 to 1 of (3x)^2-(3x^2)^2 is 6/5

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integral from 0 to 1 of (3x)^2-(3x^2)^2

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

\frac{6}{5}

Decimal Notation

1.2

Solution steps

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

Simplify

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

= \int_{0}^{1} 9 x^{2} - 9 x^{4} 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}^{1} 9 x^{2} d x - \int_{0}^{1} 9 x^{4} d x

\int_{0}^{1} 9 x^{2} d x = 3

\int_{0}^{1} 9 x^{4} d x = \frac{9}{5}

= 3 - \frac{9}{5}

Simplify

3 - \frac{9}{5} : \frac{6}{5}

= \frac{6}{5}

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\int_{0}^{\frac{7}{4}} \frac{1}{16 x ^{2} + 49} d x

What is the integral from 0 to 1 of (3x)^2-(3x^2)^2 ?
The integral from 0 to 1 of (3x)^2-(3x^2)^2 is 6/5