Let f(x) be a differentiable function defined for x > 0 . If the area bounded by the curve y = f(x) x , the x-axis, and the vertical lines x=1 and x=e^t ( t > 0 ) is given by t^2 e^t , then the value of f'(e^2) is equal to

Step-by-step solution

The area bounded by the curve is given by the definite integral: ₁^ e^t f(x) x dx = t^2 e^t Differentiating both sides with respect to t using the Newton-Leibniz formula: f(e^t) e^t d dt (e^t) = d dt (t^2 e^t) f(e^t) e^t e^t = 2t e^t + t^2 e^t f(e^t) = (t^2 + 2t) e^t Differentiating both sides with respect to t again using the chain rule: f'(e^t) e^t = (2t + 2) e^t + (t^2 + 2t) e^t Dividing by e^t (since e^t 0 ): f'(e^t) = t^2 + 4t + 2 To find f'(e^2) , we substitute t = 2 : f'(e^2) = (2)^2 + 4(2) + 2 = 4 + 8 + 2 =