AIEEE 2003 | Definite Integration Question 266 | Mathematics

If $$f\left( y \right) = {e^y},$$ $$g\left( y \right) = y;y > 0$$ and

$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $$ then :

$$F\left( t \right) = t{e^{ - t}}$$

$$F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)$$

$$F\left( t \right) = {e^t} - \left( {1 + t} \right)$$

$$F\left( t \right) = t{e^t}$$.

AIEEE 2003

MCQ (Single Correct Answer)

-1

Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} $$ is

$$e + {{{e^2}} \over 2} + {5 \over 2}$$

$$e - {{{e^2}} \over 2} - {5 \over 2}$$

$$e + {{{e^2}} \over 2} - {3 \over 2}$$

$$e - {{{e^2}} \over 2} - {3 \over 2}$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

If $$f\left( {a + b - x} \right) = f\left( x \right)$$ then $$\int\limits_a^b {xf\left( x \right)dx} $$ is equal to

$${{a + b} \over 2}\int\limits_a^b {f\left( {a + b + x} \right)dx} $$

$${{a + b} \over 2}\int\limits_a^b {f\left( {b - x} \right)dx} $$

$${{a + b} \over 2}\int\limits_a^b {f\left( x \right)dx} $$

$$\,{{b - a} \over 2}\int\limits_a^b {f\left( x \right)dx} $$