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Definite Integration question

2023 · 12 Apr · Shift 1 · Q39
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  5. /2023 · 12 Apr · Shift 1 · Q39

Definite Integration question

2023 · 12 Apr · Shift 1 · Q39

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
If ∫−0.150.15∣100x2−1∣dx=k3000\int_{-0.15}^{0.15}\left|100 x^{2}-1\right| d x=\frac{k}{3000}∫−0.150.15​​100x2−1​dx=3000k​, then kkk is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 575

  1. We need to evaluate I=∫−0.150.15∣100x2−1∣dx.I=\int_{-0.15}^{0.15}\left|100x^2-1\right|dx.I=∫−0.150.15​​100x2−1​dx.

  2. First find where the expression inside modulus changes sign: 100x2−1=0  ⟹  x2=1100  ⟹  x=±0.1.100x^2-1=0\implies x^2=\frac{1}{100}\implies x=\pm 0.1.100x2−1=0⟹x2=1001​⟹x=±0.1.

So:

  • for ∣x∣<0.1|x|<0.1∣x∣<0.1, 100x2−1<0100x^2-1<0100x2−1<0, hence ∣100x2−1∣=1−100x2;|100x^2-1|=1-100x^2;∣100x2−1∣=1−100x2;
  • for 0.1≤∣x∣≤0.150.1\le |x|\le 0.150.1≤∣x∣≤0.15, 100x2−1≥0100x^2-1\ge 0100x2−1≥0, hence ∣100x2−1∣=100x2−1.|100x^2-1|=100x^2-1.∣100x2−1∣=100x2−1.

  1. Since the integrand is even, I=2∫00.15∣100x2−1∣dx.I=2\int_0^{0.15}|100x^2-1|dx.I=2∫00.15​∣100x2−1∣dx. Thus, I=2(∫00.1(1−100x2)dx+∫0.10.15(100x2−1)dx).I=2\left(\int_0^{0.1}(1-100x^2)dx+\int_{0.1}^{0.15}(100x^2-1)dx\right).I=2(∫00.1​(1−100x2)dx+∫0.10.15​(100x2−1)dx).

  2. Compute the first integral: ∫00.1(1−100x2)dx=[x−100x33]00.1.\int_0^{0.1}(1-100x^2)dx=\left[x-\frac{100x^3}{3}\right]_0^{0.1}.∫00.1​(1−100x2)dx=[x−3100x3​]00.1​. At x=0.1x=0.1x=0.1, 0.1−100(0.1)33=0.1−100⋅0.0013=0.1−0.13=110−130=115.0.1-\frac{100(0.1)^3}{3}=0.1-\frac{100\cdot 0.001}{3}=0.1-\frac{0.1}{3}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15}.0.1−3100(0.1)3​=0.1−3100⋅0.001​=0.1−30.1​=101​−301​=151​. So, ∫00.1(1−100x2)dx=115.\int_0^{0.1}(1-100x^2)dx=\frac{1}{15}.∫00.1​(1−100x2)dx=151​.

  3. Compute the second integral: ∫0.10.15(100x2−1)dx=[100x33−x]0.10.15.\int_{0.1}^{0.15}(100x^2-1)dx=\left[\frac{100x^3}{3}-x\right]_{0.1}^{0.15}.∫0.10.15​(100x2−1)dx=[3100x3​−x]0.10.15​. Now, (100(0.15)33−0.15)−(100(0.1)33−0.1).\left(\frac{100(0.15)^3}{3}-0.15\right)-\left(\frac{100(0.1)^3}{3}-0.1\right).(3100(0.15)3​−0.15)−(3100(0.1)3​−0.1). Using fractions, 0.15=320,0.1=110.0.15=\frac{3}{20},\qquad 0.1=\frac{1}{10}.0.15=203​,0.1=101​. Then

\qquad (0.1)^3=\left(\frac{1}{10}\right)^3=\frac{1}{1000}.$$ So, $$\frac{100(0.15)^3}{3}=\frac{100}{3}\cdot\frac{27}{8000}=\frac{9}{80},$$ $$\frac{100(0.1)^3}{3}=\frac{100}{3}\cdot\frac{1}{1000}=\frac{1}{30}.$$ Hence, $$\int_{0.1}^{0.15}(100x^2-1)dx=\left(\frac{9}{80}-\frac{3}{20}\right)-\left(\frac{1}{30}-\frac{1}{10}\right).$$ Now, $$\frac{9}{80}-\frac{3}{20}=\frac{9}{80}-\frac{12}{80}=-\frac{3}{80},$$ $$\frac{1}{30}-\frac{1}{10}=\frac{1}{30}-\frac{3}{30}=-\frac{1}{15}.$$ Therefore, $$\int_{0.1}^{0.15}(100x^2-1)dx=-\frac{3}{80}+\frac{1}{15}= rac{-9+16}{240}=\frac{7}{240}.$$ 6. Add them: $$I=2\left(\frac{1}{15}+\frac{7}{240}\right)=2\left(\frac{16+7}{240}\right)=2\cdot\frac{23}{240}=\frac{23}{120}.$$ 7. Given $$I=\frac{k}{3000},$$ so $$\frac{k}{3000}=\frac{23}{120}.$$ Thus, $$k=3000\cdot\frac{23}{120}=25\cdot 23=575.$$ 8. Therefore, $$\boxed{k=575}.$$
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