Question

Let ƒ(x) be a function satisfying ƒ¢(x) = ƒ(x) with ƒ(0) = 1 and g(1) and g(x) be a function that satisfies ƒ(x) + g(x) = x². Then the value of the integral $\int _ { 0 } ^ { 1 } f$ (x) g(x) dx is -

• A. e + $\frac { \mathrm { e } ^ { 2 } } { 2 } - \frac { 3 } { 2 }$
• B. e – – $\frac { 3 } { 2 }$
• C. e + +$\frac { 5 } { 2 }$
• D. e – – $\frac { 5 } { 2 }$

Answer 1

Answer: (B)

e – – $\frac { 3 } { 2 }$

Answer 2

g(x) dx = (x² – e^x) dx

[Q ƒ(x) = e^x satisfies ƒ ¢(x) = ƒ(x) and ƒ(0) = 1]

= e^xdx –dx

= – e^x dx – $\left[ \frac { \mathrm { e } ^ { 2 \mathrm { x } } } { 2 } \right] _ { 0 } ^ { 1 }$

= e – 2$\left[ \left[ \mathrm { xe } ^ { \mathrm { x } } \right] _ { 0 } ^ { 1 } - \int _ { 0 } ^ { 1 } \mathrm { e } ^ { \mathrm { x } } \mathrm { dx } \right]$ – $\frac { 1 } { 2 }$e² + $\frac { 1 } { 2 }$

= e – 2 [e – (e – 1)] – $\frac { 1 } { 2 }$e² + $\frac { 1 } { 2 }$

= e – 2 – $\frac { 1 } { 2 }$e² + $\frac { 1 } { 2 }$ = e – $\frac { \mathrm { e } ^ { 2 } } { 2 }$ – $\frac { 3 } { 2 }$ .