Evaluate: ∫−221+5xx2dxAnswer: 38SolutionLet I=∫−221+5xx2dx. Apply ∫−aaf(x)dx=∫−aaf(−x)dx:I=∫−221+5−xx2dx=∫−221+5x5xx2dxAdd the two forms; the denominators combine to 1:2I=∫−22x2dx=2∫02x2dx=2⋅38=316Hence I=38.CBSE Class 12 Board · 2016 · Subjective