0

Guest

Courses New

Let f(x) be a function satisfying f’(x) = f(x) with f(0) = 1 and g be the function satisfying f(x) + g(x) = x 2 . The value of the integral 0 ∫ 1 f(x)g(x)dx is options e – ½ e 2 – 5/2 e – e 2 – 3 ½ (e-3) e – ½ e 2 - 3/2

Let f(x) be a function satisfying f’(x) = f(x) with f(0) = 1 and g be the function satisfying f(x) + g(x) = x2. The value of the integral 0∫1 f(x)g(x)dx is
options
  1. e – ½ e2 – 5/2
  2. e – e2 – 3
  3. ½ (e-3)
  4. e – ½ e2- 3/2

Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

f'(x) = f(x)
\frac{df(x)}{dx} = f(x)
\int \frac{df(x)}{f(x)} = \int dx
ln[f(x)] = x + c
f(x) = e^{x+c}
f(0) = e^{c}
1 = e^{c}
f(x) = e^{x}
f(x) + g(x) = x^2
e^{x} + g(x) = x^2
g(x) = x^2 - e^{x}
I = \int_{0}^{1}f(x)g(x)dx
I = \int_{0}^{1}e^{x}(x^2-e^{x})dx
I = \int_{0}^{1}x^2e^{x}dx-\int_{0}^{1}e^{2x}dx
I = (x^2e^{x}-2xe^{x}+2e^{x})_{0}^{1} - (\frac{e^{2x}}{2})_{0}^{1}
I = (e-2e+2e)_{0}^{1} - (\frac{e^{2}-1}{2})
I = (e-2e+2e-2) - (\frac{e^{2}-1}{2})
I = e-2 - \frac{e^{2}}{2}+\frac{1}{2}
I = e- \frac{e^{2}}{2}-\frac{3}{2}
Option (4)


Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More