Question

For some function f(x) and g(x) which are differentiable $\forall x > 0$ satisfy the following condition.

(i) $(\frac{f(x)}{x})' = x^2e^{-x^2}$

(ii) $g(x) = \frac{4}{e^4} \int_1^x e^{t^2} \cdot f(t) dt$

(iii) $f(1) = \frac{1}{e}$

Find the value of $3e^4(f(2) - g(2))$.

Answer 1

20

Answer 2

We are given:

1. $(\frac{f(x)}{x})' = x^2e^{-x^2}$.
2. $g(x) = \frac{4}{e^4}\int_1^x e^{t^2}f(t)\,dt$.
3. $f(1)=\frac{1}{e}$.

Step 1. Write $u(x)=\frac{f(x)}{x}$ so that $u'(x)=x^2e^{-x^2}$.

Integrate from 1 to $x$: $u(x)-u(1)=\int_1^x t^2e^{-t^2}\,dt$.

Since $u(1)=f(1)=\tfrac{1}{e}$, we get $\frac{f(x)}{x} = \frac{1}{e}+\int_1^x t^2e^{-t^2}\,dt \quad\Longrightarrow\quad f(x)=x\left(\frac{1}{e}+\int_1^x t^2e^{-t^2}\,dt\right)$.

In particular, $f(2)=2\left(\frac{1}{e}+\int_1^2 t^2e^{-t^2}\,dt\right)$.

Step 2. Express $g(2)$: $g(2)=\frac{4}{e^4}\int_1^2 e^{t^2}f(t)\,dt$.

Note that with $f(t)=t\left(\frac{1}{e}+\int_1^t s^2e^{-s^2}\,ds\right)$, we have $g(2)=\frac{4}{e^4}\int_1^2 t e^{t^2}\left(\frac{1}{e}+\int_1^t s^2e^{-s^2}\,ds\right)dt$.

Let $I=\int_1^2 t^2e^{-t^2}\,dt$, $J(t)=\int_1^t s^2e^{-s^2}\,ds$.

Then $f(2)=\frac{2}{e}+2I$, and $g(2)=\frac{4}{e^4}\left[\frac{1}{e}\int_1^2 t e^{t^2}\,dt+\int_1^2 t e^{t^2}J(t)\,dt\right]$.

Step 3. Evaluate the simpler integral: Let $A=\int_1^2 t e^{t^2}\,dt$.

Use substitution $u=t^2$ so that $du=2t\,dt$ which gives: $A=\frac{1}{2}\int_{1}^{4} e^u\,du=\frac{e^4-e}{2}$.

Thus the first term in $g(2)$ becomes: $\frac{4}{e^4}\cdot\frac{1}{e}\cdot\frac{e^4-e}{2}=\frac{4(e^4-e)}{2e^5}=\frac{2(e^4-e)}{e^5}=\frac{2}{e}-\frac{2}{e^4}$.

Step 4. Now denote $B=\int_1^2 t e^{t^2}J(t)\,dt$.

Note that $\frac{d}{dt}\Bigl(e^{t^2}J(t)\Bigr)= 2te^{t^2}J(t)+ t^2e^{-t^2}\cdot e^{t^2} =2te^{t^2}J(t)+ t^2$.

Thus, $2te^{t^2}J(t)=\frac{d}{dt}\Bigl(e^{t^2}J(t)\Bigr)-t^2$.

Integrate from 1 to 2: $2\int_1^2 t e^{t^2}J(t)\,dt = \Bigl[ e^{t^2}J(t) \Bigr]_1^2 -\int_1^2 t^2\,dt$.

At $t=1$, $J(1)=\int_1^1 s^2e^{-s^2}\,ds=0$. Also, $\int_1^2 t^2dt=\left.\frac{t^3}{3}\right|_1^2=\frac{8-1}{3}=\frac{7}{3}$.

Therefore, $2B = e^4J(2)-\frac{7}{3}\quad \Longrightarrow\quad B=\frac{e^4J(2)}{2}-\frac{7}{6}$.

But $J(2)=\int_1^2 s^2e^{-s^2}ds=I$. So, $B=\frac{e^4I}{2}-\frac{7}{6}$.

Step 5. Now write $g(2)=\left[\frac{2}{e}-\frac{2}{e^4}\right]+\frac{4}{e^4}B =\frac{2}{e}-\frac{2}{e^4}+\frac{4}{e^4}\left(\frac{e^4I}{2}-\frac{7}{6}\right) =\frac{2}{e}-\frac{2}{e^4}+2I-\frac{28}{6e^4}$.

Notice that $\frac{28}{6e^4}=\frac{14}{3e^4}$.

Thus, $g(2)=\frac{2}{e}+2I-\left(\frac{2}{e^4}+\frac{14}{3e^4}\right) =\frac{2}{e}+2I-\frac{(6+14)}{3e^4} =\frac{2}{e}+2I-\frac{20}{3e^4}$.

Recall that $f(2)=\frac{2}{e}+2I$. Hence, $f(2)-g(2)= \left(\frac{2}{e}+2I\right)-\left(\frac{2}{e}+2I-\frac{20}{3e^4}\right) =\frac{20}{3e^4}$.

Step 6. Finally, the required value is: $3e^4\Bigl(f(2)-g(2)\Bigr)=3e^4\left(\frac{20}{3e^4}\right)=20$.