Let f(x) be a function satisfying f'(x) = f(x) with f(0) = 1 and g(x) be a function that satisfies f(x) + g(x) = x^2 . Then the value of the integral - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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Let f (x) be a function satisfying f^{'} (x) = f (x) with f (0) = 1 and g (x) be a function that satisfies f (x) + g (x) = x^{2} . Then the value of the integral \int_{0}^{1} f (x) g (x) d x, is:

Solution & Step-by-step Explanation

First, find f (x) : \frac{df}{d x} = f ⟹ \int \frac{df}{f} = \int d x ⟹ ln f = x + C . f (0) = 1 ⟹ ln 1 = 0 + C ⟹ C = 0 .So f (x) = e^{x} .Then g (x) = x^{2} - e^{x} .Integral: \int_{0}^{1} e^{x} (x^{2} - e^{x}) d x = \int_{0}^{1} (x^{2} e^{x} - e^{2 x}) d x Part 1: \int x^{2} e^{x} d x = e^{x} (x^{2} - 2 x + 2) .At 0 to 1: [e (1 - 2 + 2)] - [1 (0 - 0 + 2)] = e - 2 .Part 2: \int_{0}^{1} e^{2 x} d x = [\frac{e ^{2 x}}{2}]_{0}^{1} = \frac{e ^{2}}{2} - \frac{1}{2} .Total = (e - 2) - (\frac{e ^{2}}{2} - \frac{1}{2}) = e - \frac{e ^{2}}{2} - \frac{3}{2} .

Try it yourself before checking the explanation above.

Let f (x) be a function satisfying f^{'} (x) = f (x) with f (0) = 1 and g (x) be a function that satisfies f (x) + g (x) = x^{2} . Then the value of the integral \int_{0}^{1} f (x) g (x) d x, is:

e - \frac{e ^{2}}{2} - \frac{5}{2}

e - \frac{e ^{2}}{2} + \frac{3}{2}

e - \frac{e ^{2}}{2} - \frac{3}{2}

e + \frac{e ^{2}}{2} + \frac{5}{2}

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