What is the integral from 0 to 1 of (64)/((x+2)^5) ?

The integral from 0 to 1 of (64)/((x+2)^5) is 65/81

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integral from 0 to 1 of (64)/((x+2)^5)

\int_{0}^{1} \frac{64}{( x + 2 ) ^{5}} d x

\frac{65}{81}

Decimal Notation

0.80246 \dots

Solution steps

\int_{0}^{1} \frac{64}{( x + 2 ) ^{5}} d x

Take the constant out:

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

= 64 \cdot \int_{0}^{1} \frac{1}{( x + 2 ) ^{5}} d x

Apply Integration By Parts

: [\frac{x}{( x + 2 ) ^{5}} - \int - \frac{5 x}{( x + 2 ) ^{6}} d x]_{0}^{1}

= 64 [\frac{x}{( x + 2 ) ^{5}} - \int - \frac{5 x}{( x + 2 ) ^{6}} d x]_{0}^{1}

\int - \frac{5 x}{( x + 2 ) ^{6}} d x = \frac{5}{4 ( x + 2 ) ^{4}} - \frac{2}{( x + 2 ) ^{5}}

= 64 [\frac{x}{( x + 2 ) ^{5}} - (\frac{5}{4 ( x + 2 ) ^{4}} - \frac{2}{( x + 2 ) ^{5}})]_{0}^{1}

Simplify

64 [\frac{x}{( x + 2 ) ^{5}} - (\frac{5}{4 ( x + 2 ) ^{4}} - \frac{2}{( x + 2 ) ^{5}})]_{0}^{1} : 64 [- \frac{1}{4 ( x + 2 ) ^{4}}]_{0}^{1}

= 64 [- \frac{1}{4 ( x + 2 ) ^{4}}]_{0}^{1}

Compute the boundaries:

\frac{65}{5184}

= 64 \cdot \frac{65}{5184}

Simplify

= \frac{65}{81}

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What is the integral from 0 to 1 of (64)/((x+2)^5) ?
The integral from 0 to 1 of (64)/((x+2)^5) is 65/81