∫exx2 + 1x + 12dx is equal to
exx + 1 + C
- exx + 1 + C
exx - 1x + 1 + C
exx + 1x - 1 + C
∫x + 1x1 + xexdx is equal to
log1 + xexxex + C
logxex1 + xex + C
log1 + xex + C
∫fxg'x - f'xgxfxgxloggx - logfxdx is equal to
loggxfx + C
12loggxfx2 + C
gxfxloggxfx + C
loggxfx - gxfx + C
∫0π4sinx + cosx3 + sin2xdx = ?
12log3
log2
log(3)
14log3
∫- 111 + x + x2 - 1 - x + x21 + x + x2 + 1 - x + x2dx = ?
3π2
π2
0
- 1
The area of the region described by(x, y)/x2 + y ≤ 1 and y ≤ 1 - x is
π2 - 23
π2 + 23
π2 + 43
π2 - 43
The solution of dydx + 1x = eyx2 is
2x = 1 +Cx2 ey
x = 1 +Cx2 ey
2x2 = 1 +Cx2 e - y
x2 = 1 +Cx2 e - y
The differential equation dydx = 1ax + by + c ,where a, b, c are all non-zero real numbers, is
linear in y
linear in x
linear in both x and y
homogeneous equation
Let D be the domain of a twice differentiable function f.For all x ∈ D, f"(x) + f(x) = 0 and f(x)= ∫g(x)dx + constant.If h(x) = f(x)2 + g(x)2 and h(0) = 5, then h(2015) - h(2014) is equal to
5
3
1
C.
We have,hx = fx2 + gx2Now, h'x = 2fxf'x + 2gxg'x= 2fxgx + 2gxg'x ∵ f'x = gx= 2gxfx + g'x= 2gxfx + f''x ∵ g'x = f''x= 2gx × 0 = 0 ∵ fx + f''x = 0⇒ hx = Constant = C⇒ h0 = c∴ c = 5 ∵ h0 = 5⇒ hx = 5∴ h2015 - h2014 = 5 - 5 = 0
The foci of the ellipse x216 + y2b2 = 1 and the hyperbola x2144 - y281 = 125 coincide. Then, the value of b2 is
7
9