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:

$$\left(\frac{f(x)}{x}\right)' = x^2 e^{-x^2},\quad f(1)=\frac{1}{e},\quad g(x)=\frac{4}{e^4}\int_{1}^{x}e^{t^2}f(t)\,dt.$$

1. Find f(x):

Integrate (i) from 1 to $x$:

$$\frac{f(x)}{x} - \frac{f(1)}{1} = \int_1^x t^2 e^{-t^2}\,dt.$$

Thus,

$$\frac{f(x)}{x} = \frac{1}{e} + \int_1^x t^2 e^{-t^2}\,dt \quad\Longrightarrow\quad f(x) = x\left[\frac{1}{e} + \int_1^x t^2 e^{-t^2}\,dt\right].$$

In particular,

$$f(2)= 2\left[\frac{1}{e} + I\right]\quad\text{where}\quad I=\int_1^2 t^2 e^{-t^2}\,dt.$$

2. Express g(2):

We have

$$g(2)=\frac{4}{e^4}\int_{1}^{2}e^{t^2}f(t)\,dt.$$

Write $f(t)= t\left[\frac{1}{e}+\int_1^t u^2 e^{-u^2}\,du\right]$. Then

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

Let

$$J_1 = \frac{1}{e}\int_1^2 t e^{t^2}\,dt,\quad J_2 = \int_1^2 t e^{t^2}\left(\int_1^t u^2 e^{-u^2}\,du\right)dt.$$

• Evaluating $J_1$:

Let $s=t^2$ so that $ds=2t\,dt$; then

$$\int t e^{t^2}\,dt=\frac{1}{2}e^{t^2}.$$

Thus,

$$J_1=\frac{1}{e}\cdot\frac{1}{2}\left(e^{4}-e\right)=\frac{e^{4}-e}{2e}.$$

• Evaluating $J_2$:

Change the order of integration. Write

$$J_2=\int_{t=1}^2 \left[\;t e^{t^2}\int_{u=1}^t u^2 e^{-u^2}\,du\right]dt =\int_{u=1}^{2}u^2 e^{-u^2}\left[\int_{t=u}^{2} t e^{t^2}\,dt\right]du.$$

But

$$\int_{t=u}^{2}t e^{t^2}\,dt =\frac{1}{2}\left(e^{4}-e^{u^2}\right).$$

Thus,

$$J_2=\frac{1}{2}\left[e^{4}\int_1^2u^2 e^{-u^2}\,du - \int_1^2 u^2\,du\right].$$

We note $\int_1^2 u^2 e^{-u^2}\,du = I$ (the same $I$ as before) and

$$\int_1^2 u^2\,du = \left[\frac{u^3}{3}\right]_1^2 = \frac{8-1}{3}=\frac{7}{3}.$$

Thus,

$$J_2=\frac{1}{2}\left(e^{4}I - \frac{7}{3}\right).$$

So,

$$\int_1^2 e^{t^2}f(t)\,dt = \frac{e^{4}-e}{2e} + \frac{1}{2}\left(e^{4}I - \frac{7}{3}\right).$$

Then,

$$g(2)=\frac{4}{e^4}\left[\frac{e^{4}-e}{2e} + \frac{e^{4}I}{2} - \frac{7}{6}\right] =\frac{2(e^{4}-e)}{e^5} + 2I - \frac{14}{3e^4}.$$

Notice that

$$\frac{2(e^{4}-e)}{e^5}=\frac{2}{e}-\frac{2}{e^4}.$$

Thus,

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

3. Compute $f(2)-g(2)$:

Recall $f(2)=\frac{2}{e}+2I$. Therefore,

$$\begin{aligned} f(2)-g(2) &= \left(\frac{2}{e}+2I\right) - \left[\frac{2}{e}-\frac{2}{e^4}+2I-\frac{14}{3e^4}\right]\\[1mm] &=\cancel{\frac{2}{e}}+2I-\cancel{\frac{2}{e}}+ \frac{2}{e^4}-2I+\frac{14}{3e^4}\\[1mm] &=\frac{2}{e^4}+\frac{14}{3e^4}=\frac{6+14}{3e^4}=\frac{20}{3e^4}. \end{aligned}$$

Thus,

$$3e^4(f(2)-g(2))=3e^4\cdot\frac{20}{3e^4}=20.$$