What is the integral from 0 to 1 of 5^{2t-2} ?

The integral from 0 to 1 of 5^{2t-2} is (12)/(25ln(5))

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integral from 0 to 1 of 5^{2t-2}

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

\frac{12}{25 ln ( 5 )}

Decimal

0.29824 \dots

Solution steps

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

Apply u-substitution

: \int_{- 2}^{0} 5^{u} \cdot \frac{1}{2} d u

= \int_{- 2}^{0} 5^{u} \cdot \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 2}^{0} 5^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

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

Compute the boundaries:

\frac{24}{25 ln ( 5 )}

= \frac{1}{2} \cdot \frac{24}{25 ln ( 5 )}

Simplify

= \frac{12}{25 ln ( 5 )}

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What is the integral from 0 to 1 of 5^{2t-2} ?
The integral from 0 to 1 of 5^{2t-2} is (12)/(25ln(5))