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Definite Integration question

2021 · 25 Feb · Shift 1 · Q26
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Definite Integration question

2021 · 25 Feb · Shift 1 · Q26

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of ∫−11x2e[x3]dx\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx−1∫1​x2e[x3]dx, where [ t ] denotes the greatest integer ≤\le≤ t, is :
  1. A
    e+13{{e + 1} \over 3}3e+1​
  2. B
    e−13e{{e - 1} \over {3e}}3ee−1​
  3. C
    13e{1 \over {3e}}3e1​
  4. D
    e+13e{{e + 1} \over {3e}}3ee+1​
View written solutionFree

Correct answer: D

  1. We need to evaluate I=∫−11x2e[x3] dx,I=\int_{-1}^{1} x^2 e^{[x^3]}\,dx,I=∫−11​x2e[x3]dx, where [t][t][t] denotes the greatest integer less than or equal to ttt.

  2. Observe the range of x3x^3x3 for x∈[−1,1]x\in[-1,1]x∈[−1,1]: x3∈[−1,1].x^3\in[-1,1].x3∈[−1,1]. Hence we determine [x3][x^3][x3] on subintervals:

  • For −1≤x<0-1\le x<0−1≤x<0, we have −1≤x3<0-1\le x^3<0−1≤x3<0, so [x3]=−1.[x^3]=-1.[x3]=−1.
  • At x=0x=0x=0, [0]=0[0]=0[0]=0.
  • For 0≤x<10\le x<10≤x<1, we have 0≤x3<10\le x^3<10≤x3<1, so [x3]=0.[x^3]=0.[x3]=0.
  • At x=1x=1x=1, [1]=1[1]=1[1]=1, but a single point does not affect the integral.

So, for integration purposes,

e^{-1}, & -1\le x<0,\\ 1, & 0\le x<1. \end{cases}$$ 3. Split the integral at $x=0$: $$I=\int_{-1}^{0} x^2 e^{-1}\,dx+\int_{0}^{1} x^2\,dx.$$ That is, $$I=\frac{1}{e}\int_{-1}^{0} x^2\,dx+\int_{0}^{1} x^2\,dx.$$ 4. Compute each part: $$\int x^2\,dx=\frac{x^3}{3}.$$ Therefore, $$\int_{-1}^{0} x^2\,dx=\left[\frac{x^3}{3}\right]_{-1}^{0}=0-\left(-\frac13\right)=\frac13,$$ and $$\int_{0}^{1} x^2\,dx=\left[\frac{x^3}{3}\right]_{0}^{1}=\frac13.$$ 5. Substitute back: $$I=\frac{1}{e}\cdot\frac13+\frac13=\frac{1}{3e}+\frac13.$$ Taking common denominator $3e$, $$I=\frac{1+e}{3e}.$$ 6. Compare with the options: $$\frac{e+1}{3e}$$ is option **D**. Therefore, the correct answer is **D**.
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