NIMCET 2015 — Mathematics PYQ
NIMCET | Mathematics | 2015If ∫ex(f(x)−f′(x))dx=ϕ(x), then the value of ∫exf(x)dx is:
Choose the correct answer:
- A.
ϕ(x)+exf(x)
- B.
ϕ(x)−exf(x)
- C.
21[ϕ(x)+exf(x)]
(Correct Answer) - D.
21[ϕ(x)+exf′(x)]
21[ϕ(x)+exf(x)]
Explanation
Let I=∫exf(x)dx.
Using integration by parts on I:
Take u=f(x) and v=ex.
∫uvdx=u∫vdx−∫(u′∫vdx)dx
I=f(x)ex−∫f′(x)exdx
∫exf′(x)dx=exf(x)−I…(1)
Now, the given equation is:
∫ex(f(x)−f′(x))dx=ϕ(x)
∫exf(x)dx−∫exf′(x)dx=ϕ(x)
Substitute I and equation (1) into the expression:
I−(exf(x)−I)=ϕ(x)
I−exf(x)+I=ϕ(x)
2I−exf(x)=ϕ(x)
2I=ϕ(x)+exf(x)
I=21[ϕ(x)+exf(x)]
Correct Option: (c)
Explanation
Let I=∫exf(x)dx.
Using integration by parts on I:
Take u=f(x) and v=ex.
∫uvdx=u∫vdx−∫(u′∫vdx)dx
I=f(x)ex−∫f′(x)exdx
∫exf′(x)dx=exf(x)−I…(1)
Now, the given equation is:
∫ex(f(x)−f′(x))dx=ϕ(x)
∫exf(x)dx−∫exf′(x)dx=ϕ(x)
Substitute I and equation (1) into the expression:
I−(exf(x)−I)=ϕ(x)
I−exf(x)+I=ϕ(x)
2I−exf(x)=ϕ(x)
2I=ϕ(x)+exf(x)
I=21[ϕ(x)+exf(x)]
Correct Option: (c)