What is the integral from 0 to 1 of 6x5^x ?

The integral from 0 to 1 of 6x5^x is (6(5ln(5)-4))/(ln^2(5))

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integral from 0 to 1 of 6x5^x

\int_{0}^{1} 6 x 5^{x} d x

\frac{6 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

Solution steps

\int_{0}^{1} 6 x \cdot 5^{x} d x

Take the constant out:

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

= 6 \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}

= 6 [\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 )}

= 6 [\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 )}

= 6 \cdot \frac{5 ln ( 5 ) - 4}{ln ^{2} ( 5 )}

Simplify

= \frac{6 ( 5 ln ( 5 ) - 4 )}{ln ^{2} ( 5 )}

x \to a lim (1)

2 x^{3} (2.16)

\frac{d}{d x} (120)

x \to - 0 lim (\frac{1}{x})

y = x, y = x^{2} + 1, x = - 1, x = 2

What is the integral from 0 to 1 of 6x5^x ?
The integral from 0 to 1 of 6x5^x is (6(5ln(5)-4))/(ln^2(5))