∫2xf'(x) + f(x)log2dx is
2xf'(x) + C
2xf(x) + C
2x(log(2))f(x) + C
log(2)f(x) + C
Evaluate the following integral
∫- 12xsinπxdx
I = ∫- 12xsinπxdx = ∫- 11xsinπxdx = ∫12xsinπxdx = 2∫01xsinπxdx + ∫12xsinπxdx = 2∫01xsinπxdx - ∫12xsinπxdx = 2I1 - I2 I1 = ∫01xsinπxdx = - xcosπxπ01 + ∫01cosπxπdx = - xcosπxπ + sinπxπ201 = 1πand I2 = ∫12xsinπxdx = - xcosπxπ + sinπxπ212 = - 2π + 0 + - 1π = - 3πSo, 2I1 - I2 = 2π + 3π = 5π⇒ ∫- 12xsinπxdx = 5π
∫logx3xdx is equal to
13logx2 + c
23logx2 + c
∫ex2x - 2x2dx
exx + c
ex2x2 + c
2exx + c
2exx2 + c
The value of the integral ∫dxex + e- x2
12e2x + 1 + c
12e- 2x + 1 + c
- 12e2x + 1- 1 + c
14e2x - 1 + c
∫1 + cosxdx is equal to
22cosx2 + c
22sinx2 + c
2cosx2 + c
2sinx2 + c
The value of integral ∫0π2sin5xdx is
415
85
815
45
If ddx{f(x)} = g(x), then ∫abf(x)g(x)dx is equal to
12f2(b) - f2a
12g2(b) - g2a
f(b) - f(a)
12f(b2) - fa2
If I1 = ∫03πfcos2xdx
and I2 = ∫0πfcos2xdx, then
I1 = I2
3I1 = I2
I1 = 3I2
I1 = 5I2
The value of I = ∫- π2π2sinxdx is
0
2
- 2
- 2 < 1 < 2