Integrals with Euler

Calculus Level 2

∫ − ln ⁡ 20 ln ⁡ 9 2 e x d x = ? \large \displaystyle \int _{ -\ln { 20 } }^{ \ln { 9 } }{ 2{ e }^{ x } } \, dx = \, ?

8.95 -22 -11 17.9

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Liam P by Liam P , 22, Canada

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1 solution

Chew-Seong Cheong
May 30, 2016

I = ∫ − ln ⁡ 20 ln ⁡ 9 2 e x d x = 2 e x ∣ − ln ⁡ 20 ln ⁡ 9 = 2 ( 9 − 1 20 ) = 179 10 = 17.9 \begin{aligned} I & = \int_{-\ln 20}^{\ln 9} 2e^x \ dx \\ & = 2e^x \bigg|_{-\ln 20}^{\ln 9} \\ & = 2\left(9 - \frac{1}{20} \right) \\ & = \frac{179}{10} = \boxed{17.9} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

wounderfull e^ln9= 9 ok

Patience Patience - 5 years ago

Nice solution (+1) ( ･ิϖ･ิ)

Ashish Menon - 5 years ago

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