# Integral/summation proof via induction

• Feb 5th 2010, 09:45 AM
Runty
Integral/summation proof via induction
Suppose $P$ is a polynomial of degree $k\geq 1$. Prove that

$\int P(x)e^x dx = e^x\sum_{j=0}^{k}(-1)^jP^{(j)}(x)+C$

by induction on $k$. Here, $P^{(j)}$ is the $jth$ derivative of $P$, and $P^{(0)}$ is to be interpreted as $P$.

A hint I was provided was to use integration by parts.
• Feb 5th 2010, 10:09 AM
felper
I'll let you the case $k=1$. Now, we'll suposse that the proposition is true for some poyinomial of degree $k$. We have to prove that this proposition is true for $k+1$. Let be $\underbrace{P(x)}_{k+1}$ a polynomial of degree $k+1$.

$\int\underbrace{P(x)}_{k+1}e^xdx$

Integrating by parts

$\int\underbrace{P(x)}_{k+1}(e^x)'dx=\underbrace{P( x)}_{k+1}e^x-\int\underbrace{P(x)}_{k}e^xdx$

But the proposition is true for the polynomial of degree k, then:

$\int\underbrace{P(x)}_{k+1}e^xdx = \underbrace{P(x)}_{k+1}e^x- e^x\sum_{j=0}^k(-1)^j\underbrace{P^{(j)}(x)}_{k}$

And now, conclude (Wink)