What is the integral from 0 to 1 of-2x5^x ?

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

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integral from 0 to 1 of-2x5^x

\int_{0}^{1} - 2 x 5^{x} d x

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

Decimal Notation

- 3.12489 \dots

Solution steps

\int_{0}^{1} - 2 x \cdot 5^{x} d x

Take the constant out:

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

= - 2 \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}

= - 2 [\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 )}

= - 2 [\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 )}

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

Simplify

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

3 x^{2} y^{2} d x + y d y = 5 x y

\int_{1}^{9} \frac{1}{x} d x

\int (x^{2} - 4) d x

f (x) = x^{3} - 2 x + 2, (3, 23)

\int_{0}^{x^{3}} (t + 1) (t - 2) d t

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