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Definite Integration question

2023 · 11 Apr · Shift 2 · Q35
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  5. /2023 · 11 Apr · Shift 2 · Q35

Definite Integration question

2023 · 11 Apr · Shift 2 · Q35

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let the function f:[0,2]→Rf:[0,2] \rightarrow \mathbb{R}f:[0,2]→R be defined as f(x)={emin⁡{x2,x−[x]},x∈[0,1)e[x−log⁡ex],x∈[1,2]f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}f(x)={emin{x2,x−[x]},e[x−loge​x],​x∈[0,1)x∈[1,2]​ where [t][t][t] denotes the greatest integer less than or equal to ttt. Then the value of the integral ∫02xf(x)dx\int_{0}^{2} x f(x) d x∫02​xf(x)dx is :
  1. A
    2e−12 e-12e−1
  2. B
    2e−122 e-\frac{1}{2}2e−21​
  3. C
    1+3e21+\frac{3 e}{2}1+23e​
  4. D
    (e−1)(e2+12)(e-1)\left(e^{2}+\frac{1}{2}\right)(e−1)(e2+21​)
View written solutionFree

Correct answer: B

  1. Understand the piecewise definition

The function is

f(x)={emin⁡{x2, x−[x]},x∈[0,1)e[x−ln⁡x],x∈[1,2]f(x)= \begin{cases} e^{\min\{x^2,\,x-[x]\}}, & x\in[0,1)\\[4pt] e^{[x-\ln x]}, & x\in[1,2] \end{cases}f(x)={emin{x2,x−[x]},e[x−lnx],​x∈[0,1)x∈[1,2]​

where [t][t][t] is the greatest integer function.

We need to compute

I=∫02xf(x) dxI=\int_0^2 x f(x)\,dxI=∫02​xf(x)dx

by splitting over [0,1)[0,1)[0,1) and [1,2][1,2][1,2].

  1. For x∈[0,1)x\in[0,1)x∈[0,1)

Since 0≤x<10\le x<10≤x<1, we have

[x]=0[x]=0[x]=0

so

x−[x]=x.x-[x]=x.x−[x]=x.

Thus,

min⁡{x2,x−[x]}=min⁡{x2,x}.\min\{x^2,x-[x]\}=\min\{x^2,x\}.min{x2,x−[x]}=min{x2,x}.

Now for 0≤x<10\le x<10≤x<1, we know

x2≤x,x^2\le x,x2≤x,

so

min⁡{x2,x}=x2.\min\{x^2,x\}=x^2.min{x2,x}=x2.

Hence

f(x)=ex2,x∈[0,1).f(x)=e^{x^2}, \qquad x\in[0,1).f(x)=ex2,x∈[0,1).

Therefore,

I1=∫01xex2 dx.I_1=\int_0^1 x e^{x^2}\,dx.I1​=∫01​xex2dx.

Use substitution:

u=x2  ⟹  du=2x dx.u=x^2 \implies du=2x\,dx.u=x2⟹du=2xdx.

So,

I1=12∫01eu du=12(e−1).I_1=\frac12\int_0^1 e^u\,du =\frac12(e-1).I1​=21​∫01​eudu=21​(e−1).
  1. For x∈[1,2]x\in[1,2]x∈[1,2]

We need to simplify

[x−ln⁡x].[x-\ln x].[x−lnx].

For x∈[1,2]x\in[1,2]x∈[1,2], note that

  • ln⁡x∈[0,ln⁡2]\ln x\in[0,\ln 2]lnx∈[0,ln2],
  • hence
x−ln⁡x∈[1,2].x-\ln x\in[1,2].x−lnx∈[1,2].

Let us determine its floor more carefully.

Define

g(x)=x−ln⁡x.g(x)=x-\ln x.g(x)=x−lnx.

Then

g′(x)=1−1x=x−1x≥0for x∈[1,2],g'(x)=1-\frac1x=\frac{x-1}{x}\ge 0 \quad \text{for } x\in[1,2],g′(x)=1−x1​=xx−1​≥0for x∈[1,2],

so ggg is increasing on [1,2][1,2][1,2]. Also,

g(1)=1,g(1)=1,g(1)=1, g(2)=2−ln⁡2<2.g(2)=2-\ln 2<2.g(2)=2−ln2<2.

Thus for all x∈[1,2]x\in[1,2]x∈[1,2],

1≤x−ln⁡x<2.1\le x-\ln x<2.1≤x−lnx<2.

Therefore,

[x−ln⁡x]=1.[x-\ln x]=1.[x−lnx]=1.

Hence

f(x)=e1=e,x∈[1,2].f(x)=e^1=e, \qquad x\in[1,2].f(x)=e1=e,x∈[1,2].

So,

I2=∫12xe dx=e∫12x dx=e[x22]12=e⋅4−12=3e2.I_2=\int_1^2 x e\,dx=e\int_1^2 x\,dx =e\left[\frac{x^2}{2}\right]_1^2 =e\cdot\frac{4-1}{2} =\frac{3e}{2}.I2​=∫12​xedx=e∫12​xdx=e[2x2​]12​=e⋅24−1​=23e​.
  1. Add both parts

Therefore,

I=I1+I2=12(e−1)+3e2=e−1+3e2=4e−12=2e−12.I=I_1+I_2 =\frac12(e-1)+\frac{3e}{2} =\frac{e-1+3e}{2} =\frac{4e-1}{2} =2e-\frac12.I=I1​+I2​=21​(e−1)+23e​=2e−1+3e​=24e−1​=2e−21​.
  1. Match with options

The value is

2e−12\boxed{2e-\frac12}2e−21​​

which corresponds to Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.

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