Area and Volume Formulas — Question 1

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Question 1

Assume that ff is a continuous, nonnegative function on the interval [a,b][a,b]. Prove that the area of the region bounded by the graph of y=f(x)y=f(x), the xx-axis, and the vertical lines x=ax=a and x=bx=b is A=∫abf(x)dx.A=\int_a^b f(x)\,dx.

Original worksheet page 1: question and worked solution for 7-6-001
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Question 1 - Solution

We interpret area using approximating rectangles.

Let a=x0<x1<⋯<xn=ba=x_0<x_1<\cdots<x_n=b be a partition of [a,b][a,b], and let Δxi=xi−xi−1.\Delta x_i=x_i-x_{i-1}.

Choose a sample point xi*x_i^* in each subinterval [xi−1,xi][x_{i-1},x_i].

Since f(x)≥0f(x)\ge 0, the area of the rectangle over [xi−1,xi][x_{i-1},x_i] with height f(xi*)f(x_i^*) is Areai=f(xi*)Δxi.\text{Area}_i=f(x_i^*)\,\Delta x_i.

The total area of all rectangles is ∑i=1nf(xi*)Δxi.\sum_{i=1}^n f(x_i^*)\,\Delta x_i.

As the partition is refined (that is, as ∥P∥→0\|P\|\to 0), the rectangles better approximate the region under the curve. Because ff is continuous on [a,b][a,b], it is integrable, and the limit of these sums exists.

Taking the limit gives lim∥P∥→0∑i=1nf(xi*)Δxi=∫abf(x)dx.\lim_{\|P\|\to 0}\sum_{i=1}^n f(x_i^*)\,\Delta x_i = \int_a^b f(x)\,dx.

By definition, this limit equals the exact area of the region under the curve.

Therefore, the area of the region bounded by y=f(x)y=f(x), the xx-axis, and x=ax=a, x=bx=b is A=∫abf(x)dx.A=\int_a^b f(x)\,dx.

A=∫abf(x)dx\boxed{A=\int_a^b f(x)\,dx}

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