Second Fundamental Theorem of Calculus
Trending Questions
The domain of definition of the function is given by the equation is
If denote the binomial coefficients in the expansion of
If, then is equal to ?
none of these
Find the value of
None of these
- True
- False
- π
π2
- π2−12
- π−1
- ak
- bk
- πak
- πbk
- \N
- 1
- 2
- 3
- 1∫−1t⋅f(t) dt=1011
- f(1)+f(−1)=3011
- 1∫−1t⋅f(t) dt>1∫−1f(t) dt
- f(1)−f(−1)=2011
- 1−π4
- 1+π4
- π4
- π2
- 6√3
- 4√3
- 12√3
- 2√3
∫π20 ex sin x dx= [Roorkee 1978]
- 12(ex2−1)
- 12(ex2+1)
- 12(1−ex2)
- 2(ex2+1)
- \N
- 10
- 100
- not defined
- −ka412
- ka424
- −ka424
- ka412
∫x20x+sin x1+cos xdx= [Roorkee 1978]
- -log 2
- log 2
- π2
- \N
- a−1+e2
- a+1−e2
- a+1+e2
- a+1+e22
- (−13, 2}
- (−2, 13}
- (−2, −13}
- (−13, 2)
∫π2−π2√12(1−cos 2x)dx=
0
2
- 12
None of these
- −2
- 2
- 4
- −1
The greatest value of for possible values of such that is
Let f(x) be a function satisfying f′(x)=f(x) with f(0) = 1
and g(x) be the function satisfying f(x)+g(x)=x2
The value of integral ∫10f(x)g(x) dx is equal to [AIEEE 2003; DCE 2005]
14(e−7)
- 14(e−2)
- 12(e−3)
None of these
Find the product of .
∫10e2 In xdx= [MP PET 1990]
- 0
- 12
- 13
14
Let f be a positive function. Let
I1=∫k1−k xf{x(1−x)}dx, I2=∫k1−kf{x(1−x)}dx
when 2k−1>0. Then I1I2 is [IIT 1997 Cancelled]
2
k
12
1