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Definite Integration question

2004 · Shift 0 · Q85
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Definite Integration question

2004 · Shift 0 · Q85

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of ∫−23∣1−x2∣dx\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx}−2∫3​​1−x2​dx is
  1. A
    13{1 \over 3}31​
  2. B
    143{14 \over 3}314​
  3. C
    73{7 \over 3}37​
  4. D
    283{28 \over 3}328​
View written solutionFree

Correct answer: D

  1. We need to evaluate I=∫−23∣1−x2∣ dx.I=\int_{-2}^{3}|1-x^2|\,dx.I=∫−23​∣1−x2∣dx.

  2. First, determine where the expression inside the modulus changes sign: 1−x2=0  ⟹  x=±1.1-x^2=0 \implies x=\pm 1.1−x2=0⟹x=±1.

So:

  • for ∣x∣<1|x|<1∣x∣<1, we have 1−x2≥01-x^2\ge 01−x2≥0, hence ∣1−x2∣=1−x2|1-x^2|=1-x^2∣1−x2∣=1−x2;
  • for ∣x∣>1|x|>1∣x∣>1, we have 1−x2<01-x^2<01−x2<0, hence ∣1−x2∣=x2−1|1-x^2|=x^2-1∣1−x2∣=x2−1.

  1. Split the integral at x=−1x=-1x=−1 and x=1x=1x=1: I=∫−2−1(x2−1) dx+∫−11(1−x2) dx+∫13(x2−1) dx.I=\int_{-2}^{-1}(x^2-1)\,dx+\int_{-1}^{1}(1-x^2)\,dx+\int_{1}^{3}(x^2-1)\,dx.I=∫−2−1​(x2−1)dx+∫−11​(1−x2)dx+∫13​(x2−1)dx.

  2. Compute each part.

For the first integral: ∫(x2−1)dx=x33−x.\int (x^2-1)dx=\frac{x^3}{3}-x.∫(x2−1)dx=3x3​−x. Thus,

=\left(-\frac13+1\right)-\left(-\frac83+2\right) =\frac23-\left(-\frac23\right)=\frac43.$$ For the second integral: $$\int (1-x^2)dx=x-\frac{x^3}{3}.$$ Thus, $$I_2=\left[x-\frac{x^3}{3}\right]_{-1}^{1} =\left(1-\frac13\right)-\left(-1+\frac13\right) =\frac23-\left(-\frac23\right)=\frac43.$$ For the third integral: $$I_3=\left[\frac{x^3}{3}-x\right]_{1}^{3} =\left(\frac{27}{3}-3\right)-\left(\frac13-1\right) =(9-3)-\left(-\frac23\right) =6+\frac23=\frac{20}{3}.$$ 5. Add them: $$I=I_1+I_2+I_3=\frac43+\frac43+\frac{20}{3}=\frac{28}{3}.$$ 6. Therefore, the correct option is $$\boxed{\text{D }\frac{28}{3}}.$$
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