What is the integral from 0 to 1 of x^4e^{5x} ?

The integral from 0 to 1 of x^4e^{5x} is (329e^5-24)/(3125)

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integral from 0 to 1 of x^4e^{5x}

\int_{0}^{1} x^{4} e^{5 x} d x

\frac{329 e ^{5} - 24}{3125}

Decimal Notation

15.61725 \dots

Solution steps

\int_{0}^{1} x^{4} e^{5 x} d x

Apply Integration By Parts

: [\frac{1}{5} e^{5 x} x^{4} - \int \frac{4}{5} e^{5 x} x^{3} d x]_{0}^{1}

= [\frac{1}{5} e^{5 x} x^{4} - \int \frac{4}{5} e^{5 x} x^{3} d x]_{0}^{1}

\int \frac{4}{5} e^{5 x} x^{3} d x = \frac{4}{3125} (125 e^{5 x} x^{3} - 3 (25 e^{5 x} x^{2} - 2 (5 e^{5 x} x - e^{5 x})))

= [\frac{1}{5} e^{5 x} x^{4} - \frac{4}{3125} (125 e^{5 x} x^{3} - 3 (25 e^{5 x} x^{2} - 2 (5 e^{5 x} x - e^{5 x})))]_{0}^{1}

Compute the boundaries:

\frac{329 e ^{5}}{3125} - \frac{24}{3125}

= \frac{329 e ^{5}}{3125} - \frac{24}{3125}

Simplify

= \frac{329 e ^{5} - 24}{3125}

\int a tan^{2} (θ) d θ

\frac{d}{d x} (4 x + ln (x + 1))

\frac{d y}{d x} = \frac{6 x sec ( \frac{y}{x} ) + y}{x}

x \to 0 + lim ((tan (3 x)) x)

x^{2} - 4 x + 25

What is the integral from 0 to 1 of x^4e^{5x} ?
The integral from 0 to 1 of x^4e^{5x} is (329e^5-24)/(3125)