Question:
If for a continuous function \(f(x)\), \(\int_{-\pi}^{t}(f(x)+x)\,dx=\pi^2-t^2\), for all \(t\ge-\pi\), then \(f\left(-\frac{\pi}{3}\right)\) is equal to:
If for a continuous function \(f(x)\), \(\int_{-\pi}^{t}(f(x)+x)\,dx=\pi^2-t^2\), for all \(t\ge-\pi\), then \(f\left(-\frac{\pi}{3}\right)\) is equal to:
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Differentiate both sides with respect to \(t\) (fundamental theorem of calculus) to get \(f(t)\).
Updated On: Nov 11, 2026
- \(\pi\)
- \(\frac{\pi}{2}\)
- \(\frac{\pi}{3}\)
- \(\frac{\pi}{6}\)
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The Correct Option is A
Solution and Explanation
Differentiate both sides w.r.t. \(t\): \(f(t)+t=-2t\), so \(f(t)=-3t\).
Therefore \(f\left(-\frac{\pi}{3}\right)=-3\cdot\left(-\frac{\pi}{3}\right)=\pi\).
Check: \(\int_{-\pi}^{t}(-3x+x)dx=\int_{-\pi}^{t}(-2x)dx=-(t^2-\pi^2)=\pi^2-t^2\). Correct.
Answer: (1) \(\pi\)
Therefore \(f\left(-\frac{\pi}{3}\right)=-3\cdot\left(-\frac{\pi}{3}\right)=\pi\).
Check: \(\int_{-\pi}^{t}(-3x+x)dx=\int_{-\pi}^{t}(-2x)dx=-(t^2-\pi^2)=\pi^2-t^2\). Correct.
Answer: (1) \(\pi\)
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