What is the integral from 0 to x of 9te^{3t} ?

The integral from 0 to x of 9te^{3t} is 3xe^{3x}-e^{3x}+1

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integral from 0 to x of 9te^{3t}

\int_{0}^{x} 9 t e^{3 t} d t

3 x e^{3 x} - e^{3 x} + 1

Solution steps

\int_{0}^{x} 9 t e^{3 t} d t

Take the constant out:

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

= 9 \cdot \int_{0}^{x} t e^{3 t} d t

Apply u-substitution

= 9 \cdot \int_{0}^{3 x} \frac{e ^{u} u}{9} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int_{0}^{3 x} e^{u} u d u

Apply Integration By Parts

= 9 \cdot \frac{1}{9} [e^{u} u - \int e^{u} d u]_{0}^{3 x}

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

= 9 \cdot \frac{1}{9} [e^{u} u - e^{u}]_{0}^{3 x}

Simplify

9 \cdot \frac{1}{9} [e^{u} u - e^{u}]_{0}^{3 x} : [e^{u} u - e^{u}]_{0}^{3 x}

= [e^{u} u - e^{u}]_{0}^{3 x}

Compute the boundaries:

3 x e^{3 x} - e^{3 x} + 1

= 3 x e^{3 x} - e^{3 x} + 1

What is the integral from 0 to x of 9te^{3t} ?
The integral from 0 to x of 9te^{3t} is 3xe^{3x}-e^{3x}+1