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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
If for a continuous function $f(x),$ $\int\limits_{ - \pi }^t {(f(x) + x\,\,dx)}  = {\pi ^2} - {t^2},$ for all  $t\, \ge  - \pi ,$ then $f\left( { - \frac{\pi }{3}} \right)$ is equal to
  • ✓
    $\pi $
  • B
    $\frac {\pi }{2}$
  • C
    $\frac {\pi }{3}$
  • D
    $\frac {\pi }{6}$
✓

Answer

Correct option: A.
$\pi $
a
Let $\int_{-\pi}^{t}(f(x)+x) d x=\pi^{2}-t^{2}$

$\Rightarrow \int_{-\pi}^{t} f(x) d x+\int_{-\pi}^{t} x d x=\pi^{2}-t^{2}$

$\Rightarrow \int_{-\pi}^{t} f(x) d x+\left(\frac{t^{2}}{2}-\frac{\pi^{2}}{2}\right)=\pi^{2}-t^{2}$

$\Rightarrow \int_{-\pi}^{t} f(x) d x=\frac{3}{2}\left(\pi^{2}-t^{2}\right)$

differentiating with respect to $t$

$\frac{d}{d t}\left[\int_{-\pi}^{t} f(x) d x\right]=\frac{3}{2} \frac{d}{d t}\left(\pi^{2}-t^{2}\right)$

$f(t) \cdot \frac{d t}{d t}-f(-\pi) \frac{d}{d t}(-\pi)=-3 t$

$f(t)=-3 t$

$f\left(-\frac{\pi}{3}\right)=-3\left(-\frac{\pi}{3}\right)=\pi$

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