What is the integral from 3 to 4 of 15x(x-2)^4 ?

The integral from 3 to 4 of 15x(x-2)^4 is 687/2

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integral from 3 to 4 of 15x(x-2)^4

\int_{3}^{4} 15 x (x - 2)^{4} d x

\frac{687}{2}

Decimal Notation

343.5

Solution steps

\int_{3}^{4} 15 x (x - 2)^{4} d x

Take the constant out:

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

= 15 \cdot \int_{3}^{4} x (x - 2)^{4} d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} (x - 2)^{4} - \int 2 x^{2} (x - 2)^{3} d x]_{3}^{4}

= 15 [\frac{1}{2} x^{2} (x - 2)^{4} - \int 2 x^{2} (x - 2)^{3} d x]_{3}^{4}

\int 2 x^{2} (x - 2)^{3} d x = 2 (\frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + 3 x^{4} - \frac{8 x ^{3}}{3})

= 15 [\frac{1}{2} x^{2} (x - 2)^{4} - 2 (\frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + 3 x^{4} - \frac{8 x ^{3}}{3})]_{3}^{4}

Compute the boundaries:

\frac{229}{10}

= 15 \cdot \frac{229}{10}

Simplify

= \frac{687}{2}

\int 2 x^{2} e^{- x} d x

x^{2} e^{2 x}

\int tan^{4} (x) d x

6 + \frac{2 t}{1 + t ^{3}} - e^{- 0.02 t}

\frac{d}{d s} (e^{- s})

What is the integral from 3 to 4 of 15x(x-2)^4 ?
The integral from 3 to 4 of 15x(x-2)^4 is 687/2