Use a power series to approximate the value of the integral with an error less than 0.001. Round your answer to four decimal places.
A) 0.1925
B) 0.1700
C) 0.1813
D) 0.1933
E) 0.1817

Question

Use a power series to approximate the value of the integral  with an error less than 0.001. Round your answer to four decimal places.
A) 0.1925 
B) 0.1700 
C) 0.1813 
D) 0.1933 
E) 0.1817

Quizplus · Accepted Answer

Using the power series for $e^x$, we have

$$\int_0^{0.1} e^{x^2} dx = \int_0^{0.1} \sum_{n=0}^\infty \frac{x^{2n}}{n!} dx = \sum_{n=0}^\infty \frac{0.1^{2n+1}}{(2n+1) n!} $$

We need to find an $N$ such that

$$\sum_{n=N+1}^\infty \frac{0.1^{2n+1}}{(2n+1) n!} < 0.001$$

We can estimate the remainder using the ratio test:

$$\frac{(0.1^2)^{N+1}}{(2N+3)(N+1)} < 0.001$$

Solving for $N$ using a calculator, we get $N=2$. Therefore, we need to compute the sum

$$\sum_{n=0}^2 \frac{0.1^{2n+1}}{(2n+1) n!} = 0.1813$$

Rounding to four decimal places, we get $\boxed{	extbf{(C) }0.1813}$.

Use a Power Series to Approximate the Value of the Integral