What is the integral from 0 to 1 of te^{(t^2)/2} ?

The integral from 0 to 1 of te^{(t^2)/2} is sqrt(e)-1

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integral from 0 to 1 of te^{(t^2)/2}

\int_{0}^{1} t e^{\frac{t ^{2}}{2}} d t

e - 1

Decimal

0.64872 \dots

Solution steps

\int_{0}^{1} t e^{\frac{t ^{2}}{2}} d t

Apply u-substitution

: \int_{0}^{1} \frac{e ^{\frac{u}{2}}}{2} d u

= \int_{0}^{1} \frac{e ^{\frac{u}{2}}}{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_{0}^{1} e^{\frac{u}{2}} d u

Apply u-substitution

: \int_{0}^{\frac{1}{2}} e^{v} \cdot 2 d v

= \frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} e^{v} \cdot 2 d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int_{0}^{\frac{1}{2}} e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{2} \cdot 2 [e^{v}]_{0}^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 [e^{v}]_{0}^{\frac{1}{2}} : [e^{v}]_{0}^{\frac{1}{2}}

= [e^{v}]_{0}^{\frac{1}{2}}

Compute the boundaries:

e - 1

= e - 1

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

\int_{\frac{1}{2}}^{1} 1 d x

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

\int_{0}^{1} (8 x^{3} + 3 e^{x} + 7) d x

\int_{- 1}^{1} 8 x - 2 d x

What is the integral from 0 to 1 of te^{(t^2)/2} ?
The integral from 0 to 1 of te^{(t^2)/2} is sqrt(e)-1