Q. 3 - Calculus

Question

The Calculus of Power Series: Let \(\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}\) be a power series in \(x-x_{0}\) that converges to a function \(f(x)\) on an interval \(I\).

If \(c\) and \(d\) are two points in \(I\), then the definite integral of \(f\) on the interval \([c, d ]\) is given by \(\int_{c}^{d}f\left ( x \right )dx=\) ______.

Step-by-Step Solution

Answer

If \(c\) and \(d\) are two points in \(I\), then the definite integral of \(f\) on the interval \([c, d ]\) is given by \(\begin{aligned}\int_c^d f(x) d x &=\int_c^d\left(\sum_{k=0}^{\infty} a_k\left(x-x_0\right)^k\right) d x\\&=\sum_{k=0}^{\infty} a_k \int_c^d\left(x-x_0\right)^k d x \\&=\sum_{k=0}^{\infty} \frac{a_k}{k+1}\left[\left(x-x_0\right)^{k+1}\right]_c^d\\&=\sum_{k=0}^{\infty} \frac{a_k}{k+1}\left(\left(d-x_0\right)^{k+1}-\left(c-x_0\right)^{k+1}\right)\end{aligned}\).

Step 1. Given Information

The given power series is \(\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}\) .

We have to find the definite integral of \(f\).

Step 2. Find the definite integral of \(f\)

If \(c\) and \(d\) are two points in \(I\), then the definite integral of \(f\) on the interval \([c, d ]\) is given by 

\(\begin{aligned}\int_c^d f(x) d x &=\int_c^d\left(\sum_{k=0}^{\infty} a_k\left(x-x_0\right)^k\right) d x\\&=\sum_{k=0}^{\infty} a_k \int_c^d\left(x-x_0\right)^k d x \\&=\sum_{k=0}^{\infty} \frac{a_k}{k+1}\left[\left(x-x_0\right)^{k+1}\right]_c^d\\&=\sum_{k=0}^{\infty} \frac{a_k}{k+1}\left(\left(d-x_0\right)^{k+1}-\left(c-x_0\right)^{k+1}\right)\end{aligned}\)