Let ∫−πt(f(x)+x)dx=π2−t2 ⇒∫−πtf(x)dx+∫−πtxdx=π2−t2⇒∫−πtf(x)dx+(2t2−2π2)=π2−t2⇒∫−πtf(x)dx=23(π2−t2) differentiating with respect to t dtd[∫−πtf(x)dx]=23dtd(π2−t2) f(t)⋅dtdt−f(−π)dtd(−π)=−3t f(t)=−3t f(−3π)=−3(−3π)=π
If for a continuous function f(x), ∫−πt(f(x)+xdx)=π2−t2, for all t≥−π, then f(−3π) is equal to:
Held on 12 Apr 2014 · Verified 6 Jul 2026.
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2π
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6π
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