Let t denote the greatest integer less than or equal to t . Then the value of the integral ∫ - 3 101…

Let t denote the greatest integer less than or equal to t. Then the value of the integral -3101sinπx+ecos2πxdx is equal to
  1. 521-ee
  2. 52e
  3. 522+ee
  4. 104e

Solution

Given, -3101sinπx+ecos2πxdx

Now let I1=-3101sinπxdx and I2=-3101ecos2πxdx

Now checking periodicity of sinπx we get,

0<x<1sinπx=0

1<x<2sinπx=-1

sinπx periodic with period 2

So, I1=5202sinπxdx

=52010+12-1dx=-52

Now checking periodicity of cos2πx we get,

0<x<14cos2πx=0

14<x<12cos2πx=-1

12<x<34cos2πx=-1

34<x<1cos2πx=0

So, cos2πx has periodicity of 1

So, I2=104014e0dx+1434e-1dx+341e0dx

I2=10414+1e×12+14

I2=52+52e

So, I1+I2=52e or -3101sinπx+ecos2πxdx=52e

Asked in: JEE Main 2022 (25 Jul Shift 2)

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