Definite Integrals
Trending Questions
Solve the integral I=∫R∞GMmx2dx
- GMm ln(R)
- −325
- −6085
- −6405
- −6645
- 5 ln 5+2 ln 2−3
- 5 ln 5−2 ln 2−7
- 5 ln 5−2 ln 2+3
- 5 ln 5−2 ln 2−3
- 92
- 97
- 79
- 29
The area of the triangle formed by the line and the coordinates axes is
Find the area enclosed by the curve y=x2 and the x-axis for interval x = 1 to x = 3.
9 sq. units
27.2 sq. units
10 sq. units
263 sq. units
- 2.7×1033 J
- 5.4×1033 J
- 2.7×1023 J
- 5.4×1032 J
- √5H4
- √3H4
- 2H3
- √3H2
- 13y2+4y+3+C
- (3y2+4y+3)22+C
- 3y2+4y+3+C
- (6y+4)3y2+4y+3+C
A piece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at 27ºC it weighs 30 gm. When the temperature of liquid is raised to 42ºC the metal piece weight 30.5 gm, specific gravity of the liquid at 42ºC is 1.20, then the linear expansion of the metal will be
2.316 × 10-5/ºC
4.316 × 10-5/ºC
None of these
3.316 × 10-5/ºC
Find an expression for in terms of .
- e−23
- e+23
- e−1
- e+1
- volt-second/ampere
- weber/ampere
- joule/(ampere)2
- ohm-second
If the displacement x and velocity v of a particle executing simple harmonic motion are related through the expression 4v2=25−x2, then its time period is
π
2π
4π
6π
- t3+2t+1
- t3+t+1
- t3+t2+t
- t3+t2+t+1
- 26 m
- 263 m
- 307 m
- 267 m
- 26 m
- 263 m
- 307 m
- 267 m
Using definite integral, find the area of the region between the given curve and the x-axis in the interval of [0, π].
- 1
- 2
- π2
- π
i. ∫π20(sin x + cos x)dx
ii. ∫10(3x2+4)dx
iii. ∫10(3e3x+e−x)dx
- i. 0; ii. 5; iii. e4−1e
- i. 0; ii. 5; iii. e3−1
- i. 2; ii. 5; iii. e3−1
- i. 2; ii. 5; iii. e4−1e
You are given a rod of length 'L'. The linear mass density varies as λ′. λ=a+bx. Here a & b are constants and the mass of the rod increases as x increases. Find the mass of the rod.
None of these
- 45
- 1253
- 13
- 1243
- 30 m/s
- 39 m/s
- 3 m/s
- 20 m/s
- No
- Yes
- T1<T2
- T1>T2
- T1=T2
- T1=2T2
- 13
- 1
- 0
- 12