Evaluating the derivative of $\int_0^{2x} e^{-(x^2+2xy+y^2)} dy $

Question

Math Problem Image

$\displaystyle F(x) =\int_0^{2x} e^{-(x^2+2xy+y^2)} dy$. Calculate $F'(1).$

Any idea how to solve it?

Answer 1

We have $$ F(x) = \int_0^{2x} e^{-(x+y)^2}\ dy $$

Leibniz integral rule gives

$$\begin{align} F'(x) &= \frac{d}{dx}(2x)\cdot\left.e^{-(x+y)^2}\right|_{y=2x} + \int_0^{2x} \frac{\partial}{\partial x}e^{-(x+y)^2}\ dy \\ &= 2e^{-9x^2} - \int_0^{2x} 2(x+y)e^{-(x+y)^2}\ dy \\ &= 2e^{-9x^2} + \left.e^{-(x+y)^2} \right|_{y=0}^{y=2x} \\ &= 3e^{-9x^2} - e^{-x^2} \end{align} $$

Answer 2

HINTS.

$e^{-x^2}$ can be taken out of the integral.

For the rest is a simple gaussian integral, just complete the square and solve it like a piece of cake.

Then you can evaluate $F'(1)$.

Or: fundamental theorem of calculus.

Or if you like co calculate

$$\int_0^{2x} e^{-(x+y)^2} dy$$

Call $z= x+y$

$$\int_x^{3x} e^{-z^2} dz$$

Which is straightforward in terms of the Error Function:

$$F(x) = \frac{1}{2} \sqrt{\pi } (\text{erf}(3 x)-\text{erf}(x))$$

Now

$$F'(x) = \frac{1}{2} \sqrt{\pi } \left(\frac{6 e^{-9 x^2}}{\sqrt{\pi }}-\frac{2 e^{-x^2}}{\sqrt{\pi }}\right)$$

Hence

$$F'(1) = \frac{1}{2} \left(\frac{6}{e^9 \sqrt{\pi }}-\frac{2}{e \sqrt{\pi }}\right) \sqrt{\pi } = \frac{3-e^8}{e^9}$$