Solve the problem.
-Use series to estimate the integral's value to within an error of magnitude less than $10 ^ { - 3 }$.
$$\int _ { 0 } ^ { 0.5 } \cos ^ { 2 } x d x$$
A) 0.4923
B) 0.5923
C) 0.4904
D) 0.2495

Question

Quizplus · Accepted Answer

Since we're using a series to estimate this integral, we'll need to start by finding the appropriate series. In this case, we can use the identity $$\cos^2 x = \frac{1+\cos 2x}{2}.$$ Integrating both sides from $0$ to $x$, we get $$\int_0^x \cos^2 t \, dt = \frac{1}{2} \int_0^x (1+\cos 2t) \, dt = \frac{x}{2} + \frac{1}{4} \sin 2x.$$

So, we can estimate the value of our integral by computing $\frac{1}{2} - \frac{1}{4}\sin 1$ (which is the value of the integral from $0$ to $1$, but we only need to estimate the integral from $0$ to $0.5$).

Now, we need to determine how many terms we need to include in the series to ensure our error is less than $10^{-3}$. We can do this by bounding the remainder using the formula for the remainder of a convergent alternating series: $$|R_n| = |a_{n+1}| \leq |a_{n}|$$ for $n \geq N$. In this case, our series is $$\frac{1}{4}\sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!}.$$

So, we need to find $N$ such that $$\frac{(2(0.5))^{2N+1}}{(2N+1)!} < 10^{-3}.$$ We could solve this exactly using a calculator or computer, but we can also estimate it by making a few observations.

- The denominator $(2N+1)!$ grows very quickly, so it's not likely to be the limiting factor.
- The numerator $(2(0.5))^{2N+1}$ has terms that grow very quickly at first but then start to level off. For example, we have $(2(0.5))^3 = 1$, $(2(0.5))^5 = 4$, and $(2(0.5))^7 = 16$. So, it seems reasonable to guess that $2^5 = 32$ might be a good estimate for the numerator.
- We want our error to be less than $10^{-3}$, which is much smaller than $0.5$. So, it seems reasonable to guess that we don't need to include many terms in the series.

With these observations, we can make a guess that $N = 5$ will be sufficient. Indeed, we can compute $$\frac{(2(0.5))^{11}}{11!} = \frac{1}{166320} \approx 6 \cdot 10^{-6}.$$

So, our estimate for the integral is $$\frac{1}{2} - \frac{1}{4}\sin 1 \approx 0.4904.$$

Therefore, the best choice is $\boxed{	extbf{(C)}.}$

Solve the Problem 10−310 ^ { - 3 }10−3 ∫00.5cos⁡2xdx\int _ { 0 } ^ { 0.5 } \cos ^ { 2 } x d x∫00.5​cos2xdx

Solve the Problem $10 ^ { - 3 }$ $\int _ { 0 } ^ { 0.5 } \cos ^ { 2 } x d x$