What is the integral from 1 to 3x of (x^3e^{xy}) ?

The integral from 1 to 3x of (x^3e^{xy}) is x^2(e^{3x^2}-e^x)

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integral from 1 to 3x of (x^3e^{xy})

\int_{1}^{3 x} (x^{3} e^{x y}) d y

x^{2} (e^{3 x^{2}} - e^{x})

Solution steps

\int_{1}^{3 x} x^{3} e^{x y} d y

Take the constant out:

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

= x^{3} \cdot \int_{1}^{3 x} e^{x y} d y

Apply u-substitution

: \int_{x}^{3 x^{2}} \frac{e ^{u}}{x} d u

= x^{3} \cdot \int_{x}^{3 x^{2}} \frac{e ^{u}}{x} d u

Take the constant out:

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

= x^{3} \frac{1}{x} \cdot \int_{x}^{3 x^{2}} e^{u} d u

Use the common integral:

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

= x^{3} \frac{1}{x} [e^{u}]_{x}^{3 x^{2}}

Simplify

x^{3} \frac{1}{x} [e^{u}]_{x}^{3 x^{2}} : x^{2} [e^{u}]_{x}^{3 x^{2}}

= x^{2} [e^{u}]_{x}^{3 x^{2}}

Compute the boundaries:

e^{3 x^{2}} - e^{x}

= x^{2} (e^{3 x^{2}} - e^{x})

\int_{1}^{4} 1 + \frac{1}{u} + \frac{1}{u ^{2}} d u

\int_{x}^{x^{3}} e^{t^{2}} d t

\int_{0}^{2} 4 - y^{2} d y

\int_{0}^{1} 3^{5 x} d x

\int_{0}^{6} (3 x^{2} + 1) d x

What is the integral from 1 to 3x of (x^3e^{xy}) ?
The integral from 1 to 3x of (x^3e^{xy}) is x^2(e^{3x^2}-e^x)