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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Definite Integration

NDA 2024 · Mathematics · Definite Integration

Let 3 f(x)+f((1)/(x))=(1)/(x)+1 What is 8 ∫_1^2 f(x) d x equal to?

Let 3f(x)+f(1x)=1x+13 f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}+1 What is 8∫12f(x)dx8 \int_1^2 f(x) d x equal to?

Options

  1. Aln⁡(8e)\ln (8 \sqrt{e})
  2. Bln⁡(4e)\ln (4 \sqrt{e})
  3. Cln⁡2\ln 2
  4. Dln⁡2−1\ln 2-1

Correct answer

A. ln⁡(8e)\ln (8 \sqrt{e})

Step-by-step solution

8∫12f(x)dx=8∫12(38x−x8+14)dx=∫12(3x−x+2)dx=[3ln⁡x−x22+2x]12=(3ln⁡2−42+4)−(3ln⁡1−12+2)=3ln⁡2+2−32=12+3ln⁡2=12ln⁡e+ln⁡23=ln⁡e+ln⁡8=ln⁡(8e)\begin{aligned} 8 \int_1^2 f(x) d x & =8 \int_1^2\left(\frac{3}{8 x}-\frac{x}{8}+\frac{1}{4}\right) d x \\ & =\int_1^2\left(\frac{3}{x}-x+2\right) d x \\ & =\left[3 \ln x-\frac{x^2}{2}+2 x\right]_1^2 \\ & =\left(3 \ln 2-\frac{4}{2}+4\right)-\left(3 \ln 1-\frac{1}{2}+2\right) \\ & =3 \ln 2+2-\frac{3}{2}=\frac{1}{2}+3 \ln 2 \\ & =\frac{1}{2} \ln e+\ln 2^3=\ln \sqrt{e}+\ln 8 \\ & =\ln (8 \sqrt{e})\end{aligned}
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