IBP Pish-Pash

Calculus Level pending

Compute

⌊ 100 ⋅ ∫ ln ⁡ 14 ln ⁡ 20 x e x sin ⁡ x d x ⌋ . \left\lfloor 100 \cdot \displaystyle\int_{\ln 14}^{\ln 20} x e^x \sin x \, \mathrm{d}x \right\rfloor.


The answer is 514.

Ahaan Rungta by Ahaan Rungta , 21, USA

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1 solution

Will Song
Jan 3, 2014

Hooray for bashing...

Okay so basically you just integrate by parts. Recall that the formula for this is ∫ u d v = u v − ∫ v d u \int u \, dv = uv - \int v \, du Here we let u = e x u = e^x . So what is v v ? v = ∫ x sin ⁡ x d x = sin ⁡ x − x cos ⁡ x v = \int x \sin x \, dx = \sin x - x \cos x through integration by parts so ∫ e x x sin ⁡ x d x = e x ( sin ⁡ x − x cos ⁡ x ) − ∫ e x ( sin ⁡ x − x cos ⁡ x ) d x \int e^x x \sin x \, dx = e^x ( \sin x - x \cos x) - \int e^x ( \sin x - x \cos x) \, dx We continue in this fashion, to arrive at ∫ e x x sin ⁡ x d x = e x ( x sin ⁡ x − x cos ⁡ x + cos ⁡ x ) + C \int e^x x \sin x \, dx = e^x ( x \sin x - x \cos x + \cos x) + C so the rest is very hard. We need to plug in the values and subtract. Then we get an answer of 515 − 1 515 - 1 .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Oops the result should be ∫ e x x sin ⁡ x d x = 1 2 e x ( x sin ⁡ x − x cos ⁡ x + cos ⁡ x ) + C \int e^x x \sin x \, dx = \frac{1}{2} e^x ( x \sin x - x \cos x + \cos x) + C I left out the 1 2 \dfrac{1}{2} in my haste.

Will Song - 7 years, 5 months ago

I did not get the answer after putting values.

A Brilliant Member - 7 years, 5 months ago

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