What is the integral from 0 to 1 of (5x)^2-(5x^4)^2 ?

The integral from 0 to 1 of (5x)^2-(5x^4)^2 is 50/9

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integral from 0 to 1 of (5x)^2-(5x^4)^2

\int_{0}^{1} (5 x)^{2} - (5 x^{4})^{2} d x

\frac{50}{9}

Decimal

5.55555 \dots

Solution steps

\int_{0}^{1} (5 x)^{2} - (5 x^{4})^{2} d x

Simplify

(5 x)^{2} - (5 x^{4})^{2} : 25 x^{2} - 25 x^{8}

= \int_{0}^{1} 25 x^{2} - 25 x^{8} 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} 25 x^{2} d x - \int_{0}^{1} 25 x^{8} d x

\int_{0}^{1} 25 x^{2} d x = \frac{25}{3}

\int_{0}^{1} 25 x^{8} d x = \frac{25}{9}

= \frac{25}{3} - \frac{25}{9}

Simplify

\frac{25}{3} - \frac{25}{9} : \frac{50}{9}

= \frac{50}{9}

y = (ln (x))^{2}

x \to 0 lim (2 sin (x))

\frac{\partial}{\partial x} (5 sin (\frac{5 x}{y}))

2 x^{2} + 8 x

\frac{d}{d x} (((x - 1) (2 x + 3) + 7)^{5})

What is the integral from 0 to 1 of (5x)^2-(5x^4)^2 ?
The integral from 0 to 1 of (5x)^2-(5x^4)^2 is 50/9