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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int \limits_{6}^{16} \frac{\log _{\mathrm{e}} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$ is equal to:
  • A
    $6$
  • B
    $8$
  • ✓
    $5$
  • D
    $10$
✓

Answer

Correct option: C.
$5$
c
Let $I=\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$

$I=\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}(x-22)^{2}} d x \ldots(1)$

We know

$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x)\, d x(\text { king })$

So $I=\int_{6}^{16} \frac{\log _{e}(22-x)^{2}}{\log _{e}(22-x)^{2}+\log _{e}(22-(22-x))^{2}}$

$I=\int_{0}^{16} \frac{\log _{e}(22-x)^{2}}{\log _{e} x^{2}+\log _{e}(22-x)^{2}} \,d x \ldots(2)$

$(1)+(2)$

$2 I=\int_{6}^{16} 1 .\, d x=10$

$I=5$

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