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

The integral from-1 to 1 of (1-4x)^2 is 38/3

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

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

\frac{38}{3}

Decimal

12.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{5}^{- 3} - \frac{u ^{2}}{4} d u

= \int_{5}^{- 3} - \frac{u ^{2}}{4} d u

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

= - \int_{- 3}^{5} - \frac{u ^{2}}{4} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{- 3}^{5} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- 3}^{5}

= - (- \frac{1}{4} [\frac{u ^{3}}{3}]_{- 3}^{5})

Simplify

= \frac{1}{4} [\frac{u ^{3}}{3}]_{- 3}^{5}

Compute the boundaries:

\frac{152}{3}

= \frac{1}{4} \cdot \frac{152}{3}

Simplify

= \frac{38}{3}

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What is the integral from-1 to 1 of (1-4x)^2 ?
The integral from-1 to 1 of (1-4x)^2 is 38/3