```                   the value of integrand limit [0 to [x]] {x-[x]}dx (where [.] denotes greatest integer function.
```

2 years ago

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```                                        $\int_{0}^{[x]}x-[x]dx \\=>[x]^2/2-\int_{1}^{2}1dx-\int_{2}^{3}2dx-\int_{3}^{4}3dx..............depends\, value \,of\, [x]$Arun KumarIIT DelhiAskiitians Faculty
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5 months ago
```                                        x-[x] is periodic with period 1. therefore, ?0[x]x-[x] dx = [x]?01x-[x] dx =[x]/2we can get integral of x-[x] within limits 0 and 1 using method of areas as ½.Therefore the answer is [x]/2
```
5 months ago

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Q . THE AREA ENCLOSED BETWEEN THE CURVES - y = ln ( x + e ) and x = ln ( 1/y ) AND X-AXIS.

Ans: Hello Student, Please find answer to your question below Intersect x-axis at (1-e, 0) & y-axis at (0, 1) Intersect y-axis at (0, 1). Area A b/w two curves is:

 Jitender Singh 8 days ago

thanks sir......and happy diwali.......

 bharat makkar 8 days ago
see attachment and explain it

 Jitender Singh 3 months ago
1∫ -1 ( x + 3) dx

integration of x+3 from -1 to 1 is equal to (((x^2)/2)+3x) from -1 to 1 ((1/2)+(3))-((1/2)-3) =3+3 =6

 Miryala Gopalakrishna 4 days ago

Ans:- 1 ∫ -1 (x+3)dx =[(x 2 /2)+3x] -1 1 =[(1 2 )/2 -(-1 2 )/2]+3(1-(-1)) =(1/2)-(1/2)+6 =6

 anusha 4 days ago
Is the topic Vectors hard?

very easy.....make a firm grip on it becauz it is the basic thing of phy PLZ APPROVE MY ANS

 larisha sharma one year ago

if u wont command it then u will face a lot of problems in many chapters...

 Saurabh Anand one year ago

it is not very tough but you can take it as average.