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

The integral from 0 to 3 of (3x)^2-(x^2)^2 is 162/5

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

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

\frac{162}{5}

Decimal

32.4

Solution steps

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

Simplify

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

= \int_{0}^{3} 9 x^{2} - 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}^{3} 9 x^{2} d x - \int_{0}^{3} x^{4} d x

\int_{0}^{3} 9 x^{2} d x = 81

\int_{0}^{3} x^{4} d x = \frac{243}{5}

= 81 - \frac{243}{5}

Simplify

81 - \frac{243}{5} : \frac{162}{5}

= \frac{162}{5}

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

\int_{- 2}^{5} 5 d x

\int_{- 3}^{2} b x - a x^{2} d x

\int_{0}^{\infty} x^{2} e^{\frac{- x}{2}} d x

\int_{0}^{1} e^{- ik x} d x

What is the integral from 0 to 3 of (3x)^2-(x^2)^2 ?
The integral from 0 to 3 of (3x)^2-(x^2)^2 is 162/5