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

The integral from 0 to 1 of pi(4-4x)^2 is (16pi)/3

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

\int_{0}^{1} π (4 - 4 x)^{2} d x

\frac{16 π}{3}

Decimal

16.75516 \dots

Solution steps

\int_{0}^{1} π (4 - 4 x)^{2} d x

Take the constant out:

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

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

Apply u-substitution

: \int_{4}^{0} - \frac{u ^{2}}{4} d u

= π \cdot \int_{4}^{0} - \frac{u ^{2}}{4} d u

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

= π (- \int_{0}^{4} - \frac{u ^{2}}{4} d u)

Take the constant out:

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

= π (- (- \frac{1}{4} \cdot \int_{0}^{4} u^{2} d u))

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{4}

= π (- (- \frac{1}{4} [\frac{u ^{3}}{3}]_{0}^{4}))

Simplify

π (- (- \frac{1}{4} [\frac{u ^{3}}{3}]_{0}^{4})) : \frac{π}{4} [\frac{u ^{3}}{3}]_{0}^{4}

= \frac{π}{4} [\frac{u ^{3}}{3}]_{0}^{4}

Compute the boundaries:

\frac{64}{3}

= \frac{π}{4} \cdot \frac{64}{3}

Simplify

= \frac{16 π}{3}

\int_{0}^{9} 2 x + 9 d x

\int_{0}^{1} x^{2} (2 x^{3} + 3)^{2} d x

\int_{0}^{4} 1 + x d x

\int_{1}^{2} \frac{t}{t ^{2} - 1} d t

\int_{1}^{2} 1

What is the integral from 0 to 1 of pi(4-4x)^2 ?
The integral from 0 to 1 of pi(4-4x)^2 is (16pi)/3