Preparation for the Calc BC exam

Calculus Level 3

∫ ∫ 1 ∞ z − 2 d z ∫ 0 1 ln ⁡ y d y 1 − cos ⁡ 2 ( π x ) d x \large \displaystyle \int \limits_{ \large \int_{1}^{\infty} z^{-2} \, dz }^{ \large \int_{0}^{1} \ln y \, dy } \sqrt{1-\cos^{2} (\pi x)} \, dx .

4 π \dfrac{4}{\pi} The integral is of an indeterminate form. − 4 π - \dfrac{4}{\pi} 0 + ∞ + \infty − ∞ - \infty

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Rishabh Jain
May 3, 2016

Upper limit: [ y ( ln ⁡ y − 1 ) ] 0 1 = − 1 [y(\ln y-1)]_0^1=-1 .

(For substituting 0 use Lhopital( since form is 0/0) i.e lim ⁡ y → 0 ln ⁡ y − 1 1 / y = lim ⁡ y → 0 1 / y − 1 / y 2 = 0 \small{\displaystyle\lim_{y\to 0}\dfrac{\ln y-1}{1/y}=\displaystyle\lim_{y\to 0}\dfrac{1/y}{-1/y^2}=0} .)

Lower limit: [ − 1 / z ] 1 ∞ = 1 [-1/z]_1^{^{\infty}}=1 .

Also using 1 − cos ⁡ 2 A = ∣ sin ⁡ A ∣ \sqrt{1-\cos^2 A}=|\sin A| integration is:

− ∫ − 1 1 ∣ sin ⁡ π x ∣ = − 2 ∫ 0 1 sin ⁡ π x = 2 π [ c o s π x ] 0 1 = − 4 π -\int_{-1}^1|\sin \pi x|=-2\int_0^1\sin \pi x=\frac{2}{\pi}[cos \pi x]_0^1=\dfrac{-4}{\pi}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

You should show how you get the upper limit. You do need L'Hopital's rule for that because otherwise, what's 0 ln ⁡ 0 0 \ln 0 ?

Otherwise, great solution!

Hobart Pao - 5 years, 1 month ago

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I've added it

Rishabh Jain - 5 years, 1 month ago

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Thanks! Upvoted.

Hobart Pao - 5 years, 1 month ago

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