Reciprocals

Calculus Level 3

( ∫ − 2 1 ( x 2 + 4 x + 13 ) d x ) ( ∫ − 2 1 1 x 2 + 4 x + 13 d x ) \large \left( \int_{-2}^1 (x^2 + 4x+13) \, dx \right) \left( \int_{-2}^1 \dfrac1{x^2+4x+13} \, dx\right)

If the value of the expression above is equal to A π A \pi , find A A A^A .


The answer is 27.

Joel Yip by Joel Yip , 21, Singapore

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1 solution

Rishabh Jain
Feb 25, 2016

( ∫ − 2 1 ( x 2 + 4 x + 13 ) d x ) = x 3 3 + 2 x 2 + 13 x ∣ − 2 1 = 36 \large \left( \int_{-2}^1 (x^2 + 4x+13) \, dx \right)\\ \large =\dfrac{x^3}{3}+2x^2+13x|_{-2}^1=\color{#D61F06}{36} ( ∫ − 2 1 1 x 2 + 4 x + 13 d x ) = ( ∫ − 2 1 1 ( x + 2 ) 2 + 9 d x ) = 1 3 ( tan ⁡ − 1 ( x + 2 3 ) ) ∣ − 2 1 = π 12 \left( \int_{-2}^1 \dfrac1{x^2+4x+13} \, dx\right)\\ \large=\left( \int_{-2}^1 \dfrac1{(x+2)^2+9} \, dx\right) \\ \large=\dfrac{1}{3}(\tan^{-1} (\frac{x+2}{3}))|_{-2}^1=\color{#D61F06}{\dfrac{\pi}{12}} ( 36 ) × ( π 12 ) = 3 π \huge \color{#D61F06}{(36)\times(\dfrac{\pi}{12})=3\pi} ∴ 3 3 = 27 \Large \therefore~ 3^3=\huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{27}}}}}

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How did you get the third line of second integration?

I am new to integration and I presently in class 9.Can you give some sources where to learn from?Your solutions are awesome.You are a big inspiration for me..Thanks

Anik Mandal - 5 years, 3 months ago

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This should help Link

Joel Yip - 5 years, 3 months ago

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Thanks @Joel Yip

Anik Mandal - 5 years, 3 months ago

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