What is the integral from 0 to 1 of 7x5^x ?

The integral from 0 to 1 of 7x5^x is (7(5ln(5)-4))/(ln^2(5))

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integral from 0 to 1 of 7x5^x

\int_{0}^{1} 7 x 5^{x} d x

\frac{7 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

Decimal Notation

10.93712 \dots

Solution steps

\int_{0}^{1} 7 x \cdot 5^{x} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} x \cdot 5^{x} d x

Apply Integration By Parts

: [\frac{5 ^{x} x}{ln ( 5 )} - \int \frac{5 ^{x}}{ln ( 5 )} d x]_{0}^{1}

= 7 [\frac{5 ^{x} x}{ln ( 5 )} - \int \frac{5 ^{x}}{ln ( 5 )} d x]_{0}^{1}

\int \frac{5 ^{x}}{ln ( 5 )} d x = \frac{5 ^{x}}{ln ^{2} ( 5 )}

= 7 [\frac{5 ^{x} x}{ln ( 5 )} - \frac{5 ^{x}}{ln ^{2} ( 5 )}]_{0}^{1}

Compute the boundaries:

\frac{5 ln ( 5 ) - 4}{ln ^{2} ( 5 )}

= 7 \cdot \frac{5 ln ( 5 ) - 4}{ln ^{2} ( 5 )}

Simplify

= \frac{7 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

\frac{\partial}{\partial y} (x^{2} + x^{2} + y)

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

\frac{( 2 x + 1 )}{x + 2}

\int \frac{x - 1}{x ^{2} - x - 2} d x

\frac{d}{d t} (2 t^{4})

What is the integral from 0 to 1 of 7x5^x ?
The integral from 0 to 1 of 7x5^x is (7(5ln(5)-4))/(ln^2(5))