Not defined Really!

Calculus Level 3

If I 1 = ∫ 1 2 2 x 2012 − 1 x 2014 + 1 d x \displaystyle { I }_{ 1 } = \int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { { x }^{ 2012 } - 1 }{ { x }^{ 2014 } + 1 }dx } and I 2 = ∫ 2 4 ( log ⁡ x 2 − ( log ⁡ x 2 ) 2 ln ⁡ 2 ) d x \displaystyle { I }_{ 2 } = \int _{ 2 }^{ 4 }{ \left( \log _{ x }{ 2 } - \frac { { \left( \log _{ x }{ 2 } \right) }^{ 2 } }{ \ln { 2 } } \right) dx } , find the value of I 1 + I 2 { I }_{ 1 } + { I }_{ { 2 } } .

None of the above 0 ∞ \infty not defined

Priyanshu Mishra by Priyanshu Mishra , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Jul 23, 2017

I 1 = ∫ 1 2 2 x 2012 − 1 x 2014 + 1 d x = ∫ 1 2 1 x 2012 − 1 x 2014 + 1 d x + ∫ 1 2 x 2012 − 1 x 2014 + 1 d x Let u = 1 x ⟹ d u = − 1 x 2 d x = ∫ 1 2 1 x 2012 − 1 x 2014 + 1 d x + ∫ 1 1 2 1 u 2012 − 1 1 u 2014 + 1 ( − 1 u 2 ) d u = ∫ 1 2 1 x 2012 − 1 x 2014 + 1 d x − ∫ 1 2 1 u 2012 − 1 u 2014 + 1 d u = 0 \begin{aligned} I_1 & = \int_\frac 12^2 \frac {x^{2012}-1}{x^{2014}+1} dx \\ & = \int_\frac 12^1 \frac {x^{2012}-1}{x^{2014}+1} dx + \color{#3D99F6} \int_1^2 \frac {x^{2012}-1}{x^{2014}+1} dx & \small \color{#3D99F6} \text{Let } u = \frac 1x \implies du = - \frac 1{x^2} \ dx \\ & = \int_\frac 12^1 \frac {x^{2012}-1}{x^{2014}+1} dx + \color{#3D99F6} \int_1^\frac 12 \frac {\frac 1{u^{2012}}-1}{\frac 1{u^{2014}}+1}\left(-\frac 1{u^2}\right) du \\ & = \int_\frac 12^1 \frac {x^{2012}-1}{x^{2014}+1} dx - \int_\frac 12^1 \frac {u^{2012}-1}{u^{2014}+1} du \\ & = 0 \end{aligned}

I 2 = ∫ 2 4 ( log ⁡ x 2 − log ⁡ x 2 2 ln ⁡ 2 ) d x = ∫ 2 4 ( ln ⁡ 2 ln ⁡ x − ln ⁡ 2 ln ⁡ 2 x ) d x = ln ⁡ 2 ∫ 2 4 d d x ( x ln ⁡ x ) d x = ln ⁡ 2 [ x ln ⁡ x ] 2 4 = ln ⁡ 2 ( 4 2 ln ⁡ 2 − 2 ln ⁡ 2 ) = 0 \begin{aligned} I_2 & = \int_2^4 \left(\log_x 2 - \frac {\log_x^2 2}{\ln 2}\right) dx \\ & = \int_2^4 \left(\frac {\ln 2}{\ln x} - \frac {\ln 2}{\ln^2 x}\right) dx \\ & = \ln 2 \int_2^4 \frac d{dx} \left(\frac x{\ln x}\right) dx \\ & = \ln 2 \left[\frac x{\ln x}\right]_2^4 \\ & = \ln 2 \left(\frac 4{2\ln 2} - \frac 2{\ln 2} \right) \\ & = 0 \end{aligned}

⟹ I 1 + I 2 = 0 + 0 = 0 \implies I_1+I_2 = 0 + 0 = \boxed{0}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you please tell the motto behind splitting the integral between two intervals?

Priyanshu Mishra - 3 years, 10 months ago

Log in to reply

Not the limits of integral 2 and its reciprocal 1 2 \frac 12 . The integrand is 0 when x = 1 x=1 . I have seen the graph, before x = 1 x=1 , it is negative area and after it, it is positive. Then I realized function such as f ( x ) = x − 1 x f(x) = x - \frac 1x will have the similar effect, therefore, taking the reciprocal. Like I said very good question.

Chew-Seong Cheong - 3 years, 10 months ago

Log in to reply

OKAY. Now i undesrtood the basic idea/reason behind splitting integrals. Thanks a lot.

Can you please tell me a good Calculus book which has difficult questions?

Priyanshu Mishra - 3 years, 10 months ago

Log in to reply

@Priyanshu Mishra – No, I don't read any book. I am a retired person who has not been in teaching.

Chew-Seong Cheong - 3 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...