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

The integral from 0 to 4 of (x^2-4x+4)^2 is 64/5

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

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

\frac{64}{5}

Decimal Notation

12.8

Solution steps

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

Complete the square

x^{2} - 4 x + 4 : (x - 2)^{2}

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{64}{5}

= \frac{64}{5}

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What is the integral from 0 to 4 of (x^2-4x+4)^2 ?
The integral from 0 to 4 of (x^2-4x+4)^2 is 64/5