If for a continuous function f(x), ∫−πt(f(x)+x)dx=π2−t2, for all t≥-π, then f-π3 is equal to

∫−πt(f(x)+x)dx=π2−t2Differentiate w.r.t 't'f(t)+t=-2t (by Leibnitz rule) f(t)=-3tf-π3=-3-π3=π

If for a continuous function f(x), ∫−πt(f(x)+x)dx=π2−t2, for all t≥-π, then f-π3 is equal to

∫−πt(f(x)+x)dx=π2−t2Differentiate w.r.t 't'f(t)+t=-2t (by Leibnitz rule) f(t)=-3tf-π3=-3-π3=π

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If for a continuous function f(x), $\int_{- π}^{t} (f (x) + x) d x = π^{2} - t^{2},$ for all $t \geq - π,$ then $f (- \frac{π}{3})$ is equal to

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$π$

$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{6}$

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Detailed Solution

$\int_{- π}^{t} (f (x) + x) dx = π^{2} - t^{2}$

Differentiate w.r.t 't'

$(f (t) + t) = - 2 t$

(by Leibnitz rule)

$\begin{array}{rcl} f (t) & = & - 3 t \\ f (\frac{- π}{3}) & = & - 3 \frac{(- π)}{3} = π \end{array}$

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If for a continuous function f(x), ∫−πt(f(x)+x)dx=π2−t2, for all t≥-π, then f-π3 is equal to

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If for a continuous function f(x), $\int_{- π}^{t} (f (x) + x) d x = π^{2} - t^{2},$ for all $t \geq - π,$ then $f (- \frac{π}{3})$ is equal to