The function $g$ is defined by $g(x)=\int_{-20}^x (f(t))^2dt,$ and we're given $$g(20)=100,f(0)=4,f'(0)=12, f''(0)=20.$$ I am to use a tangent line approximation to estimate $g(0.1)$.
I believe I have a solution but I am not using $f'$ nor $f''.$
By linear approximation $g(0.1)\approx g(0)+.1g'(0).$
Since $f^2$ is even, $g(0)= 1/2 \cdot g(20)=50.$ By the fundamental theorem, $g'(0)=(f(0))^2=16$. This gives $g(0.1)\approx 51.6$.
Is there a solution that makes use of $f', f''$?
Best Answer
Write the MacLaurin series up to the $3$rd degree
$$g(x)=\int_{-20}^0 f(t)^2 \, dt+f(0)^2 x+f(0) x^2 f'(0)+\frac{1}{3} x^3 \left(f(0) f''(0)+f'(0)^2\right)+O(x^4)$$ Thus we have $$g(0.1)\approx\int_{-20}^0 f(t)^2 \, dt+\frac{f(0)^2}{10}+\frac{1}{100} f(0) f'(0)+\frac{f(0) f''(0)+f'(0)^2}{3000}$$ Knowing that $g(20)=100$ implies that $\int_{-20}^0 f(t)^2 \, dt=50$
Furthermore we know $f(0)=4,f'(0)=12, f''(0)=20$. So we get
$$g(0.1)\approx 50+\frac{4^2}{10}+\frac{1}{100}\,(4\cdot 12)+\frac{4\cdot 20+12^2}{3000}=\frac{19558}{375}\approx 52.1547$$
Hope this helps