What is the integral from 1 to 2 of pi(e^y)^2 ?

The integral from 1 to 2 of pi(e^y)^2 is (pi(e^4-e^2))/2

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integral from 1 to 2 of pi(e^y)^2

\int_{1}^{2} π (e^{y})^{2} d y

\frac{π ( e ^{4} - e ^{2} )}{2}

Decimal

74.15587 \dots

Solution steps

\int_{1}^{2} π (e^{y})^{2} d y

Take the constant out:

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

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

Apply u-substitution

: \int_{e}^{e^{2}} u d u

= π \cdot \int_{e}^{e^{2}} u d u

Apply the Power Rule

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

= π [\frac{u ^{2}}{2}]_{e}^{e^{2}}

Compute the boundaries:

\frac{e ^{4} - e ^{2}}{2}

= π \frac{e ^{4} - e ^{2}}{2}

Simplify

= \frac{π ( e ^{4} - e ^{2} )}{2}

\int_{- 2}^{3} e^{x + 2} d x

\int_{0}^{1} (x^{2} + 5 x) d x

\int_{0}^{2} e^{∣ 1 - x ∣} d x

\int_{0}^{1} \frac{1}{x ^{2} + x + 1} d x

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

What is the integral from 1 to 2 of pi(e^y)^2 ?
The integral from 1 to 2 of pi(e^y)^2 is (pi(e^4-e^2))/2