What is the integral from 1 to 2 of xy^3e^{x^2y^2} ?

The integral from 1 to 2 of xy^3e^{x^2y^2} is (y(e^{4y^2}-e^{y^2}))/2

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integral from 1 to 2 of xy^3e^{x^2y^2}

\int_{1}^{2} x y^{3} e^{x^{2} y^{2}} d x

\frac{y ( e ^{4 y^{2}} - e ^{y^{2}} )}{2}

Solution steps

\int_{1}^{2} x y^{3} e^{x^{2} y^{2}} d x

Take the constant out:

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

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

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

y^{2} x^{2} = (y x)^{2}

= y^{3} \cdot \int_{1}^{2} e^{(y x)^{2}} x d x

Apply u-substitution

: \int_{y}^{y \cdot 2} \frac{e ^{u^{2}} u}{y ^{2}} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int_{y^{2}}^{4 y^{2}} \frac{e ^{v}}{2} d v

= y^{3} \frac{1}{y ^{2}} \cdot \int_{y^{2}}^{4 y^{2}} \frac{e ^{v}}{2} d v

Take the constant out:

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

= y^{3} \frac{1}{y ^{2}} \cdot \frac{1}{2} \cdot \int_{y^{2}}^{4 y^{2}} e^{v} d v

Use the common integral:

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

= y^{3} \frac{1}{y ^{2}} \cdot \frac{1}{2} [e^{v}]_{y^{2}}^{4 y^{2}}

Simplify

y^{3} \frac{1}{y ^{2}} \cdot \frac{1}{2} [e^{v}]_{y^{2}}^{4 y^{2}} : \frac{y [ e ^{v} ] _{y^{2}}^{4 y^{2}}}{2}

= \frac{y [ e ^{v} ] _{y^{2}}^{4 y^{2}}}{2}

Compute the boundaries:

e^{4 y^{2}} - e^{y^{2}}

= \frac{y ( e ^{4 y^{2}} - e ^{y^{2}} )}{2}

x y^{^{'}} - 2 y = 9 x^{2}

y = 3 x^{3} + 5 x + 4, (2, 38)

\frac{d y}{d x} = y x + y^{2}

\int \frac{x ^{2} + 5 x}{( x - 3 ) ( x ^{2} + 9 )} d x

y = x^{2} - 37, y = 11 - 2 x

What is the integral from 1 to 2 of xy^3e^{x^2y^2} ?
The integral from 1 to 2 of xy^3e^{x^2y^2} is (y(e^{4y^2}-e^{y^2}))/2