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Definite Integration question

2021 · 26 Feb · Shift 2 · Q31
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  5. /2021 · 26 Feb · Shift 2 · Q31

Definite Integration question

2021 · 26 Feb · Shift 2 · Q31

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f(x)=∫0xetf(t)dt+exf(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}}f(x)=0∫x​etf(t)dt+ex be a differentiable function for all x ∈\in∈ R. Then f(x) equals :
  1. A
    e(ex−1){e^{({e^{x - 1}})}}e(ex−1)
  2. B
    2eex−12{e^{{e^x}}} - 12eex−1
  3. C
    2eex−1−12{e^{{e^x} - 1}} - 12eex−1−1
  4. D
    eex−1{e^{{e^x}}} - 1eex−1
View written solutionFree

Correct answer: C

  1. Given integral equation

We are given

f(x)=∫0xetf(t) dt+ex. f(x)=\int_0^x e^t f(t)\,dt + e^x.f(x)=∫0x​etf(t)dt+ex.

We need to determine f(x)f(x)f(x).

  1. Differentiate both sides

Since fff is differentiable, differentiate with respect to xxx:

ddxf(x)=ddx(∫0xetf(t) dt)+ddx(ex).\frac{d}{dx}f(x)=\frac{d}{dx}\left(\int_0^x e^t f(t)\,dt\right)+\frac{d}{dx}(e^x).dxd​f(x)=dxd​(∫0x​etf(t)dt)+dxd​(ex).

By the Fundamental Theorem of Calculus,

ddx(∫0xetf(t) dt)=exf(x).\frac{d}{dx}\left(\int_0^x e^t f(t)\,dt\right)=e^x f(x).dxd​(∫0x​etf(t)dt)=exf(x).

Also,

ddx(ex)=ex.\frac{d}{dx}(e^x)=e^x.dxd​(ex)=ex.

So,

f′(x)=exf(x)+ex=ex(f(x)+1).f'(x)=e^x f(x)+e^x=e^x(f(x)+1).f′(x)=exf(x)+ex=ex(f(x)+1).

Thus the differential equation is

f′(x)=ex(f(x)+1).f'(x)=e^x(f(x)+1).f′(x)=ex(f(x)+1).
  1. Find the initial condition

Put x=0x=0x=0 in the original equation:

f(0)=∫00etf(t) dt+e0=0+1=1.f(0)=\int_0^0 e^t f(t)\,dt + e^0 = 0+1=1.f(0)=∫00​etf(t)dt+e0=0+1=1.

So,

f(0)=1.f(0)=1.f(0)=1.
  1. Solve the differential equation

Let

y=f(x)+1.y=f(x)+1.y=f(x)+1.

Then

y′=f′(x).y'=f'(x).y′=f′(x).

Using the differential equation,

y′=ex(f(x)+1)=exy.y'=e^x(f(x)+1)=e^x y.y′=ex(f(x)+1)=exy.

So,

dydx=exy.\frac{dy}{dx}=e^x y.dxdy​=exy.

Separate variables:

dyy=exdx.\frac{dy}{y}=e^x dx.ydy​=exdx.

Integrate:

∫1y dy=∫ex dx\int \frac{1}{y}\,dy = \int e^x\,dx∫y1​dy=∫exdx ln⁡∣y∣=ex+C.\ln |y|=e^x+C.ln∣y∣=ex+C.

Hence,

y=Ceex.y=Ce^{e^x}.y=Ceex.

Therefore,

f(x)+1=Ceex.f(x)+1=Ce^{e^x}.f(x)+1=Ceex.

So,

f(x)=Ceex−1.f(x)=Ce^{e^x}-1.f(x)=Ceex−1.
  1. Use the initial condition

Since f(0)=1f(0)=1f(0)=1,

1=Cee0−1=Ce−1.1=Ce^{e^0}-1=Ce-1.1=Cee0−1=Ce−1.

Thus,

Ce=2⇒C=2e.Ce=2 \quad \Rightarrow \quad C=\frac{2}{e}.Ce=2⇒C=e2​.

Therefore,

f(x)=2eeex−1=2eex−1−1.f(x)=\frac{2}{e}e^{e^x}-1=2e^{e^x-1}-1.f(x)=e2​eex−1=2eex−1−1.
  1. Match with the options

We obtained

f(x)=2eex−1−1.\boxed{f(x)=2e^{e^x-1}-1}.f(x)=2eex−1−1​.

This matches Option C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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