What is the integral from 0 to 1 of 8x^7(6-x^8) ?

The integral from 0 to 1 of 8x^7(6-x^8) is 11/2

Popular >

integral from 0 to 1 of 8x^7(6-x^8)

\int_{0}^{1} 8 x^{7} (6 - x^{8}) d x

\frac{11}{2}

Decimal

5.5

Solution steps

\int_{0}^{1} 8 x^{7} (6 - x^{8}) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} x^{7} (6 - x^{8}) d x

Expand

x^{7} (6 - x^{8}) : 6 x^{7} - x^{15}

= 8 \cdot \int_{0}^{1} 6 x^{7} - x^{15} 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

= 8 (\int_{0}^{1} 6 x^{7} d x - \int_{0}^{1} x^{15} d x)

\int_{0}^{1} 6 x^{7} d x = \frac{3}{4}

\int_{0}^{1} x^{15} d x = \frac{1}{16}

= 8 (\frac{3}{4} - \frac{1}{16})

Simplify

8 (\frac{3}{4} - \frac{1}{16}) : \frac{11}{2}

= \frac{11}{2}

\int_{- 3}^{4} (1 - x) d x

\int_{0}^{1} e^{y} - 1 d y

\int_{0}^{1} cos (x^{4}) d x

\int_{- 6}^{- 3} (u - \frac{3}{u ^{2}}) d u

\int_{0}^{\frac{L}{2}} \frac{2 x}{L} sin (\frac{nπ x}{L}) d x

What is the integral from 0 to 1 of 8x^7(6-x^8) ?
The integral from 0 to 1 of 8x^7(6-x^8) is 11/2