What is the integral from 0 to 1 of t13^{t^2} ?

The integral from 0 to 1 of t13^{t^2} is 6/(ln(13))

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integral from 0 to 1 of t13^{t^2}

\int_{0}^{1} t 1 3^{t^{2}} d t

\frac{6}{ln ( 13 )}

Decimal

2.33922 \dots

Solution steps

\int_{0}^{1} t \cdot 1 3^{t^{2}} d t

Apply u-substitution

: \int_{1}^{13} \frac{1}{2 ln ( 13 )} d u

= \int_{1}^{13} \frac{1}{2 ln ( 13 )} d u

Integral of a constant:

\int a d x = a x

= [\frac{1}{2 ln ( 13 )} u]_{1}^{13}

Compute the boundaries:

\frac{6}{ln ( 13 )}

= \frac{6}{ln ( 13 )}

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What is the integral from 0 to 1 of t13^{t^2} ?
The integral from 0 to 1 of t13^{t^2} is 6/(ln(13))