Approximating Definite Integrals — Question 4

Question 4

For increasing $f$, explain why $$L_n\le\int_a^bf(x)\,dx\le R_n.$$ Then bound $\int_0^1e^x\,dx$ using $n=2$.

Question 4 – Solution

Step 1: Justify the inequality. On each subinterval, an increasing function satisfies $$f(x_{i-1})\le f(x)\le f(x_i).$$ Multiplying by $\Delta x>0$ and adding over all subintervals gives $$L_n\le\int_a^bf(x)\,dx\le R_n.$$

Step 2: Find the width and nodes. $$\Delta x=\frac{1-0}{2}=\frac12,\qquad x_0=0,\ x_1=\frac12,\ x_2=1.$$ Step 3: Compute both sums. $$\begin{align*} L_2&=\frac12[f(0)+f(1/2)]=\frac12(1+e^{1/2})\approx1.3244,\\ R_2&=\frac12[f(1/2)+f(1)]=\frac12(e^{1/2}+e)\approx2.1835. \end{align*}$$ Step 4: State the bound. $$\boxed{1.3244\le\int_0^1e^xdx\le2.1835}$$