Approximating Definite Integrals — Question 4

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Question 4

For increasing ff, explain why Ln≤∫abf(x)dx≤Rn.L_n\le\int_a^bf(x)\,dx\le R_n. Then bound ∫01exdx\int_0^1e^x\,dx using n=2n=2.

Original worksheet page 1: question and worked solution for 1-10-004
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Question 4 – Solution

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

Step 2: Find the width and nodes. Δx=1−02=12,x0=0,x1=12,x2=1.\Delta x=\frac{1-0}{2}=\frac12,\qquad x_0=0,\ x_1=\frac12,\ x_2=1. Step 3: Compute both sums. L2=12[f(0)+f(1/2)]=12(1+e1/2)≈1.3244,R2=12[f(1/2)+f(1)]=12(e1/2+e)≈2.1835.\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. 1.3244≤∫01exdx≤2.1835\boxed{1.3244\le\int_0^1e^xdx\le2.1835}

Original worksheet page 2: question and worked solution for 1-10-004

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