Using the four non-zero terms of Maclaurin Series how can I find the value of
$$\int_0^1 2x \cos ^2 x dx$$
When I solved I got Maclaurin Series as $2x-2x^3+\frac{2}{3}x^5$.
But I don't know how to apply the intervals.
Can anyone show how to do this.
Best Answer
First note that the first 4 (four) terms are $$2 x-2 x^3+\frac{2}{3} x^5 - \frac{4}{45} x^7$$ and the integral from this approximation is $\frac{3}{5}$. This is not too bad compared to the exact value $$\cos(1) \sin(1)+\frac{1}{2}\cos(1)^2 \approx 0.600612004276\approx 0.6 = \frac{3}{5}$$
Edit: An antiderivative of the four term expression is $$F(x) = x^2-\frac{1}{2}x^4+\frac{1}{9}x^6-\frac{1}{90}x^8$$ and therefore you get the integal as $F(1)-F(0) = \frac{3}{5} - 0$.