What is the integral from 0 to 1 of (5^x-2^x) ?

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

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integral from 0 to 1 of (5^x-2^x)

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

\frac{4}{ln ( 5 )} - \frac{1}{ln ( 2 )}

Decimal

1.04264 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} 5^{x} d x = \frac{4}{ln ( 5 )}

\int_{0}^{1} 2^{x} d x = \frac{1}{ln ( 2 )}

= \frac{4}{ln ( 5 )} - \frac{1}{ln ( 2 )}

\int_{0}^{3} 2 x (x - 6) d x

\int_{- 1}^{0} 2 x^{2} (4 x^{3} - 1)^{3} d x

\int_{0}^{2} 2 x e^{x} - e^{x} + e^{- x} d x

\int_{0}^{4} x^{2} e^{x} d x

\int_{- 4}^{4} - e^{(\frac{1}{4}) x} d x

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