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Area Under the Curves question

2025 · 4 Apr · Shift 1 · Q35
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  5. /2025 · 4 Apr · Shift 1 · Q35

Area Under the Curves question

2025 · 4 Apr · Shift 1 · Q35

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
Let f:[0,∞)→Rf:[0, \infty) \rightarrow \mathbb{R}f:[0,∞)→R be a differentiable function such that f(x)=1−2x+∫0xex−tf(t)dtf(x)=1-2 x+\int_0^x e^{x-t} f(t) d tf(x)=1−2x+∫0x​ex−tf(t)dt for all x∈[0,∞)x \in[0, \infty)x∈[0,∞). Then the area of the region bounded by y=f(x)y=f(x)y=f(x) and the coordinate axes is
  1. A
    5\sqrt55​
  2. B
    2
  3. C
    2\sqrt22​
  4. D
    12\frac{1}{2}21​
View written solutionFree

Correct answer: D

  1. Given integral equation

We have

f(x)=1−2x+∫0xex−tf(t) dt,x≥0.f(x)=1-2x+\int_0^x e^{x-t}f(t)\,dt, \qquad x\ge 0.f(x)=1−2x+∫0x​ex−tf(t)dt,x≥0.

We need the area of the region bounded by y=f(x)y=f(x)y=f(x) and the coordinate axes.

  1. Differentiate the equation

Let

I(x)=∫0xex−tf(t) dt.I(x)=\int_0^x e^{x-t}f(t)\,dt.I(x)=∫0x​ex−tf(t)dt.

Then by Leibniz rule,

I′(x)=ex−xf(x)+∫0x∂∂x(ex−tf(t)) dt=f(x)+∫0xex−tf(t) dt=f(x)+I(x).I'(x)=e^{x-x}f(x)+\int_0^x \frac{\partial}{\partial x}\big(e^{x-t}f(t)\big)\,dt =f(x)+\int_0^x e^{x-t}f(t)\,dt =f(x)+I(x).I′(x)=ex−xf(x)+∫0x​∂x∂​(ex−tf(t))dt=f(x)+∫0x​ex−tf(t)dt=f(x)+I(x).

Now from the given equation,

f(x)=1−2x+I(x).f(x)=1-2x+I(x).f(x)=1−2x+I(x).

Differentiating,

f′(x)=−2+I′(x)=−2+f(x)+I(x).f'(x)=-2+I'(x)=-2+f(x)+I(x).f′(x)=−2+I′(x)=−2+f(x)+I(x).

But I(x)=f(x)−1+2xI(x)=f(x)-1+2xI(x)=f(x)−1+2x, so

f′(x)=−2+f(x)+f(x)−1+2x=2f(x)+2x−3.f'(x)=-2+f(x)+f(x)-1+2x=2f(x)+2x-3.f′(x)=−2+f(x)+f(x)−1+2x=2f(x)+2x−3.

Thus,

f′(x)−2f(x)=2x−3.f'(x)-2f(x)=2x-3.f′(x)−2f(x)=2x−3.
  1. Find the initial condition

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

f(0)=1−0+∫00ex−tf(t) dt=1.f(0)=1-0+\int_0^0 e^{x-t}f(t)\,dt=1.f(0)=1−0+∫00​ex−tf(t)dt=1.

So,

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

We solve

f′(x)−2f(x)=2x−3.f'(x)-2f(x)=2x-3.f′(x)−2f(x)=2x−3.

Try a particular solution of the form

fp=ax+b.f_p=ax+b.fp​=ax+b.

Then

fp′=a.f_p'=a.fp′​=a.

Substitute:

a−2(ax+b)=2x−3.a-2(ax+b)=2x-3.a−2(ax+b)=2x−3.

Comparing coefficients,

−2a=2  ⟹  a=−1,-2a=2 \implies a=-1,−2a=2⟹a=−1,

and

a−2b=−3  ⟹  −1−2b=−3  ⟹  b=1.a-2b=-3 \implies -1-2b=-3 \implies b=1.a−2b=−3⟹−1−2b=−3⟹b=1.

So,

fp=1−x.f_p=1-x.fp​=1−x.

Homogeneous solution:

fh=Ce2x.f_h=Ce^{2x}.fh​=Ce2x.

Hence,

f(x)=Ce2x+1−x.f(x)=Ce^{2x}+1-x.f(x)=Ce2x+1−x.

Using f(0)=1f(0)=1f(0)=1,

C+1=1  ⟹  C=0.C+1=1 \implies C=0.C+1=1⟹C=0.

Therefore,

f(x)=1−x.\boxed{f(x)=1-x}.f(x)=1−x​.
  1. Find the bounded region with coordinate axes

The curve is

y=1−x.y=1-x.y=1−x.

It meets the axes at:

  • yyy-axis: (0,1)(0,1)(0,1)
  • xxx-axis: (1,0)(1,0)(1,0)

So the bounded region is the right triangle formed by the axes and the line x+y=1x+y=1x+y=1.

Area =

12×1×1=12.\frac{1}{2}\times 1 \times 1 = \frac{1}{2}.21​×1×1=21​.

Thus, the required area is

12.\boxed{\frac{1}{2}}.21​​.
  1. Check with options

Option D is

12.\boxed{\frac{1}{2}}.21​​.

So the correct option is D.

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