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

The integral from 0 to 1 of 5(1-y)^4 is 1

Popular >

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

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

1

Solution steps

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

Take the constant out:

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

= 5 \cdot \int_{0}^{1} (1 - y)^{4} d y

Apply u-substitution

: \int_{1}^{0} - u^{4} d u

= 5 \cdot \int_{1}^{0} - u^{4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 5 (- \int_{0}^{1} - u^{4} d u)

Take the constant out:

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

= 5 (- (- \int_{0}^{1} u^{4} d u))

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{0}^{1}

= 5 (- (- [\frac{u ^{5}}{5}]_{0}^{1}))

Simplify

= 5 [\frac{u ^{5}}{5}]_{0}^{1}

Compute the boundaries:

\frac{1}{5}

= 5 \cdot \frac{1}{5}

Simplify

= 1

\int_{1}^{\infty} \frac{5}{x ^{2} + 9} d x

\int_{0}^{x} \frac{cos ( t ) - 1}{t} d t

\int_{3}^{6} \frac{2}{x - 2} d x

\int_{- 1}^{2} (2 x^{2} - x^{3}) d x

\int_{1}^{e^{2}} 2 π y ln (y) d y

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