A calculus problem by Hamza A

Calculus Level 3

∫ 0 1 x 2 e x sin ⁡ x d x \large \int_0^1 x^2 e^x \sin x \, dx

If the integral above is equal to a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 3.

Hamza A by Hamza A , 19, USA

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1 solution

I = ∫ 0 1 x 2 e x sin ⁡ x d x = ∫ 0 1 x 2 e x ⋅ ℑ ( e i x ) d x = ℑ ∫ 0 1 x 2 e ( 1 + i ) x d x = ℑ ∫ 0 1 1 ( 1 + i ) 2 ⋅ ∂ 2 ∂ 2 a e a ( 1 + i ) x d x Introduce a temporary variable a = 1 = ℑ [ 1 ( 1 + i ) 2 ⋅ ∂ 2 ∂ 2 a ∫ 0 1 e a ( 1 + i ) x d x ] = ℑ [ 1 ( 1 + i ) 2 ⋅ ∂ 2 ∂ 2 a ( e a ( 1 + i ) − 1 a ( 1 + i ) ) ] = ℑ [ 1 ( 1 + i ) 2 ( d z d a ) 2 d 2 d 2 z ( e z − 1 z ) ] = ℑ [ 1 ( 1 + i ) 2 ( 1 + i ) 2 d d z ( e z z − e z − 1 z 2 ) ] = ℑ [ e z z − e z z 2 − e z z 2 + 2 ( e z − 1 ) z 3 ] = ℑ [ ( z 2 − 2 z + 2 ) e z − 2 z 3 ] Putting back a = 1 ⇒ z = 1 + i = ℑ [ ( 2 i − 2 − 2 i + 2 ) e z − 2 2 i ( 1 + i ) ] = ℑ [ 1 1 − i ] = ℑ [ 1 + i 2 ] = 1 2 \begin{aligned} I & = \int_0^1 x^2e^x \sin x \space dx \\ & = \int_0^1 x^2e^x \cdot{} \Im \left(e^{ix}\right) \space dx \\ & = \Im \int_0^1 x^2 e^{(1+i)x} \space dx \\ & = \Im \int_0^1 \frac{1}{(1+i)^2} \cdot{} \frac{\partial^2}{\partial^2 \color{#3D99F6}{a}} e^{\color{#3D99F6}{a}(1+i)x} \space dx \quad \quad \small \color{#3D99F6}{\text{Introduce a temporary variable }a = 1} \\ & = \Im \left[ \frac{1}{(1+i)^2} \cdot{} \frac{\partial^2}{\partial^2 a} \int_0^1 e^{a(1+i)x} \space dx \right] \\ & = \Im \left[ \frac{1}{(1+i)^2} \cdot{} \frac{\partial^2}{\partial^2 a} \left( \frac{ e^{a(1+i)} - 1}{a(1+i)}\right) \right] \\ & = \Im \left[ \frac{1}{(1+i)^2} \left(\frac{dz}{da}\right)^2 \frac{d^2}{d^2 z} \left( \frac{ e^z - 1}{z}\right) \right] \\ & = \Im \left[ \frac{1}{(1+i)^2} \left(1+i \right)^2 \frac{d}{dz} \left( \frac{ e^z}{z} - \frac{e^z - 1}{z^2} \right) \right] \\ & = \Im \left[ \frac{e^z}{z} - \frac{e^z}{z^2} - \frac{e^z}{z^2} + \frac{2(e^z-1)}{z^3} \right] \\ & = \Im \left[ \frac{(z^2 - 2z + 2)e^z-2}{z^3} \right] \quad \quad \small \color{#3D99F6}{\text{Putting back }a = 1 \space \Rightarrow z = 1+i} \\ & = \Im \left[ \frac{(2i - 2-2i + 2)e^z-2}{2i(1+i)} \right] \\ & = \Im \left[ \frac{1}{1-i} \right] = \Im \left[ \frac{1+i}{2} \right] = \frac{1}{2} \end{aligned}

⇒ a + b = 1 + 2 = 3 \Rightarrow a+b = 1+2 = \boxed{3} .

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Sir, can you show how does the partial derivative from the 6th row transform to (dz/da)^2 and d2/dz2 in the 7th row. Does it involved chain rule?

boon yang - 5 years, 2 months ago

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Let f ( a ) = e a ( 1 + i ) − 1 a ( 1 + i ) f(a) = \dfrac{e^{a(1+i)}-1}{a(1+i)} , then d d x f ( a ) = d d z d z d x f ( a ) \dfrac{d}{dx} f(a) = \dfrac{d}{dz} \dfrac{dz}{dx} f(a) . Please note that d z d x = 1 + i \dfrac{dz}{dx} = 1+i is a constant, therefore, d d x f ( a ) = d z d x d d z f ( a ) \dfrac{d}{dx} f(a) = \dfrac{dz}{dx} \dfrac{d}{dz} f(a) and d 2 d x 2 f ( a ) = d z d x d d z d z d x ( d d z f ( a ) ) = ( d z d x ) 2 d 2 d z 2 f ( a ) \dfrac{d^2}{dx^2} f(a) = \dfrac{dz}{dx} \dfrac{d}{dz} \dfrac{dz}{dx} \left( \dfrac{d}{dz} f(a) \right) = \left(\dfrac{dz}{dx}\right)^2 \dfrac{d^2}{dz^2} f(a)

Chew-Seong Cheong - 5 years, 2 months ago

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