Proof of Various Integral Properties — Question 8

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Question 8

Assume that ff is integrable on [a,b][a,b] and that cc satisfies a<c<b.a<c<b. Prove that ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx.\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx.

Original worksheet page 1: question and worked solution for 7-5-008
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Question 8 - Solution

We prove this property using Riemann sums.

Let a=x0<x1<⋯<xk=c<⋯<xn=ba=x_0<x_1<\cdots<x_k=c<\cdots<x_n=b be a partition of [a,b][a,b] that includes the point cc.

Let Δxi=xi−xi−1\Delta x_i = x_i-x_{i-1} and choose a sample point xi*x_i^* in each subinterval [xi−1,xi][x_{i-1},x_i].

A Riemann sum for ff over [a,b][a,b] is ∑i=1nf(xi*)Δxi.\sum_{i=1}^n f(x_i^*)\,\Delta x_i.

Split this sum at cc: ∑i=1nf(xi*)Δxi=∑i=1kf(xi*)Δxi+∑i=k+1nf(xi*)Δxi.\sum_{i=1}^n f(x_i^*)\,\Delta x_i = \sum_{i=1}^k f(x_i^*)\,\Delta x_i + \sum_{i=k+1}^n f(x_i^*)\,\Delta x_i.

The first sum is a Riemann sum for ff over [a,c][a,c], and the second sum is a Riemann sum for ff over [c,b][c,b].

Now take the limit as the norm of the partition ∥P∥→0\|P\|\to 0.

Since ff is integrable on [a,b][a,b], it is integrable on both subintervals [a,c][a,c] and [c,b][c,b].

Thus, lim∥P∥→0∑i=1kf(xi*)Δxi=∫acf(x)dx,\lim_{\|P\|\to 0}\sum_{i=1}^k f(x_i^*)\,\Delta x_i = \int_a^c f(x)\,dx, lim∥P∥→0∑i=k+1nf(xi*)Δxi=∫cbf(x)dx.\lim_{\|P\|\to 0}\sum_{i=k+1}^n f(x_i^*)\,\Delta x_i = \int_c^b f(x)\,dx.

Therefore, ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx.\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx.

∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx\boxed{\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx}

Original worksheet page 2: question and worked solution for 7-5-008

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