Triple Integral Attack

Calculus Level 5

A = ( 2015 x 2015 + 2014 ) d x + ( 2015 x 2015 + 2014 ) d x B = 0 ( 2015 x 2015 + 2015 ) d x + 0 ( 2015 x 2015 + 2015 ) d x C = 0 ( 2015 x 2015 + 2016 ) d x + 0 ( 2015 x 2015 + 2016 ) d x \begin{aligned} \displaystyle & A&=\int _{ -\infty }^{ \infty }{ ({ 2 }015x^{ 2015 }+2014) } dx+\int _{ \infty }^{ -\infty }{ ({ 2 }015x^{ 2015 }+2014)dx } \\ \displaystyle &B&=\int _{ 0 }^{ \infty }{ ({ 2 }015x^{ 2015 }+2015) } dx+\int _{ \infty }^{ 0 }{ ({ 2 }015x^{ 2015 }+2015)dx } \\\displaystyle & C&= \int _{ 0 }^{ -\infty }{ ({ 2 }015x^{ 2015 }+2016) } dx+\int _{ -\infty }^{ 0 }{ ({ 2 }015x^{ 2015 }+2016)dx } \end{aligned}

Let A , B , C A,B,C denote the values of the first, second, third expression as stated above respectively.

Evaluate A + B + C A + B + C .

Indeterminate \infty 12090 6045 0 -12090 -\infty -6045

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1 solution

Param Prashar
May 30, 2015

Because it is from infinity to minus infinity

No, it is because we are adding divergent integrals. Integrals with infinite bounds are called improper integrals .

Zach Abueg - 3 years, 10 months ago

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