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

The integral from 1 to 2 of (2x-4)^4 is 16/5

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

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

\frac{16}{5}

Decimal

3.2

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 2}^{0}

= \frac{1}{2} [\frac{u ^{5}}{5}]_{- 2}^{0}

Compute the boundaries:

\frac{32}{5}

= \frac{1}{2} \cdot \frac{32}{5}

Simplify

= \frac{16}{5}

y^{^{'}} - \frac{2}{t} y = t^{2} e^{t}, y (1) = 3

5^{x^{2}}

(1 - x^{3}) y^{^{'}} - 2 (1 + x) y = y^{\frac{5}{2}}

\int 8 + 6 x + 24 x^{2} d x

\frac{d}{d x} (\frac{9 e ^{x} x ^{\frac{7}{2}}}{2} + e^{x} x^{\frac{9}{2}})

What is the integral from 1 to 2 of (2x-4)^4 ?
The integral from 1 to 2 of (2x-4)^4 is 16/5