What is the integral from 0 to 1 of (1-x^7)^2 ?

The integral from 0 to 1 of (1-x^7)^2 is 49/60

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integral from 0 to 1 of (1-x^7)^2

\int_{0}^{1} (1 - x^{7})^{2} d x

\frac{49}{60}

Solution steps

\int_{0}^{1} (1 - x^{7})^{2} d x

Expand

(1 - x^{7})^{2} : 1 - 2 x^{7} + x^{14}

= \int_{0}^{1} 1 - 2 x^{7} + x^{14} 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} 1 d x - \int_{0}^{1} 2 x^{7} d x + \int_{0}^{1} x^{14} d x

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 2 x^{7} d x = \frac{1}{4}

\int_{0}^{1} x^{14} d x = \frac{1}{15}

= 1 - \frac{1}{4} + \frac{1}{15}

Simplify

1 - \frac{1}{4} + \frac{1}{15} : \frac{49}{60}

= \frac{49}{60}

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y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 18, y^{^{'}} (0) = \frac{31}{4}

\frac{d}{d x} (2 cos (x) + 3 sin (x))

\frac{s + 3}{s ^{2} + 2 s + 2}

5 x^{2} + 8 x - 8

What is the integral from 0 to 1 of (1-x^7)^2 ?
The integral from 0 to 1 of (1-x^7)^2 is 49/60