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

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

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

\int_{0}^{1} π (x - x^{4})^{2} d x

\frac{π}{9}

Decimal Notation

0.34906 \dots

Solution steps

\int_{0}^{1} π (x - x^{4})^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{1} (x - x^{4})^{2} d x

Factor

x - x^{4} : - (- x + x^{4})

= π \cdot \int_{0}^{1} (- (- x + x^{4}))^{2} d x

Simplify

(- (- x + x^{4}))^{2} : (- x + x^{4})^{2}

= π \cdot \int_{0}^{1} (- x + x^{4})^{2} d x

Expand

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

= π \cdot \int_{0}^{1} x^{2} - 2 x^{5} + 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

= π (\int_{0}^{1} x^{2} d x - \int_{0}^{1} 2 x^{5} d x + \int_{0}^{1} x^{8} d x)

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

\int_{0}^{1} 2 x^{5} d x = \frac{1}{3}

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

= π (\frac{1}{3} - \frac{1}{3} + \frac{1}{9})

Simplify

π (\frac{1}{3} - \frac{1}{3} + \frac{1}{9}) : \frac{π}{9}

= \frac{π}{9}

e^{t + 7}

\int 5 x 1 - x^{4} d x

x (x^{2} + 4)^{5}

f (x) = 5 x + 2 x^{2} + 7 e^{x}

y = - x^{2} + 1, y = 0

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