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

The integral from 0 to 1 of (2-x)^4 is 31/5

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

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

\frac{31}{5}

Decimal Notation

6.2

Solution steps

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

Apply u-substitution

: \int_{2}^{1} - u^{4} d u

= \int_{2}^{1} - u^{4} d u

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

= - \int_{1}^{2} - u^{4} d u

Take the constant out:

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

= - (- \int_{1}^{2} u^{4} d u)

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{2}

= - (- [\frac{u ^{5}}{5}]_{1}^{2})

Simplify

= [\frac{u ^{5}}{5}]_{1}^{2}

Compute the boundaries:

\frac{31}{5}

= \frac{31}{5}

\int_{- 1}^{1} (x^{3} - x^{5}) d x

\int_{2}^{3} \frac{2 x}{1 + x ^{2}} d x

\int_{0}^{3} e^{\frac{x}{3}} (x^{2} + 2 x) d x

\int_{- 2}^{2} \frac{8}{4 + x ^{2}} d x

\int_{2}^{4} (x^{2} + 2 x) d x

What is the integral from 0 to 1 of (2-x)^4 ?
The integral from 0 to 1 of (2-x)^4 is 31/5