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Differentiating an Integral

April 10th, 2009 |

Author: Merrin

The following is a very useful technique in calculus, taking the derivative of an integral. Leibnitz’s rule for differentiating an integral is

$\displaystyle \cfrac{d}{dx} \int_{\phi (x)}^{\chi (x)} f(x,t) dt = \chi'(x)f(x,\chi(x))-\phi'(x)f(x,\phi(x)) +\int_{\phi (x)}^{\chi (x)} \cfrac{\partial}{\partial x} f(x,t) dt$

Probably the most straightforward way to differentiate an integral is to calculate the integral and then differentiate that function. Sometimes the integral can’t be done in terms of elementary functions but by using Leibniz’s formula you can find an exact expression for the derivative of an integral. We now know all the pieces of Leibniz’s formula: ordinary derivatives, partial derivatives, and integration so we can try to apply this formula to some problems.

Example 1. Find the derivative of

$\displaystyle I(x) = \int_{e^x}^{x^2} \sin tdt$

Solution 1. This integral is trivial, and we could first calculate the integral and then differentiate but let us try Leibniz’s formula for the derivative of an integral.

$\displaystyle (x^2)'\sin (x^2) - (e^x)'\sin(e^x) + \int_{e^x}^{x^2} \cfrac{\partial}{\partial x}\sin tdt$

The regular derivatives of $latex x^2 &s=1$ and $latex e^x &s=1$ are simple and the partial derivative term is just zero because the integrand is independent of $latex x&s=1$.

$I'(x) = 2x \sin (x^2) - e^x\sin(e^x)$

Example 2. Find the derivative of

$\displaystyle I(x) = \int_x^{x^2} \cfrac{\sin(tx)} {t} dt$

Solution 2. This integral can’t be solved in terms of elementary functions, so for this problem our only option is to use Leibniz’s formula for the derivative of an integral. The partial derivative of the integrand will cancel the nasty factor of $latex t&s=1$ in the denominator and give us something we can integrate.

$\displaystyle I'(x) = (2x)\cfrac{\sin(x^3)} {x^2} - (1)\cfrac{\sin(x^2)} {x}+ \int_x^{x^2}\cfrac{\partial}{\partial x} \cfrac{\sin(tx)}{t} dt$

$\displaystyle I'(x) = \cfrac{2}{x}\sin(x^3) - \cfrac{1}{x}\sin(x^2) + \int_x^{x^2} \cos(tx) dt$

Now we have something we can integrate. Since

$\displaystyle \int_x^{x^2} \cos(tx) dt = \sin(x^3)/x-\sin(x^2)/x$

We can combine these terms with the others and we find

$I'(x) = \cfrac{3} {x} \sin(x^3) - \cfrac{2} {x}\sin(x^2)$

Sometimes differentiating an integral makes a new integral solvable. We will revisit this technique when we study how differentiation under the integral sign can be used as an advanced integration technique.

Example 3. Find the following derivative

$\displaystyle \cfrac{d}{dx} \int_0^2e^xdx$

Solution 3. Here is where you can start to get into all sorts of trouble. The variable in the integrand $latex x&s=1$ should not have anything to do with the $latex x&s=1$ of the derivative. In fact, you can replace $latex x&s=1$ in the integrand by $latex t&s=1$ or something else and you should get the same answer. If Leibniz’s formula for the derivative of an integral were applied then you would get a contribution from the partial derivative as the way the symbols stand. That would be incorrect. If you have $latex x&s=1$ in your integrand to make it a function of $latex x&s=1$ you should be careful and know what you are doing. Clearly since the limits of integration are not variable the derivative of this integral should be zero, once you clean up the dummy variable $latex x&s=1$ and replace it with something else.

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