Proof of Various Integral Properties — Question 5

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Question 5

Assume that ff is integrable on [a,b][a,b] and that f(x)≥0for all x∈[a,b].f(x)\ge 0 \quad\text{for all }x\in[a,b]. Prove that ∫abf(x)dx≥0.\int_a^b f(x)\,dx \ge 0.

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Question 5 - Solution

We use the definition of the definite integral in terms of Riemann sums.

Let a=x0<x1<⋯<xn=ba=x_0<x_1<\cdots<x_n=b be a partition of the interval [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].

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

Since f(x)≥0f(x)\ge 0 for all x∈[a,b]x\in[a,b] and Δxi>0\Delta x_i>0, each term in the sum satisfies f(xi*)Δxi≥0.f(x_i^*)\,\Delta x_i \ge 0.

Thus, every Riemann sum for ff over [a,b][a,b] is nonnegative: ∑i=1nf(xi*)Δxi≥0.\sum_{i=1}^n f(x_i^*)\,\Delta x_i \ge 0.

Now take the limit as the norm of the partition ∥P∥→0\|P\|\to 0. Since ff is integrable, the limit of the Riemann sums exists and equals the definite integral: ∫abf(x)dx=lim∥P∥→0∑i=1nf(xi*)Δxi.\int_a^b f(x)\,dx = \lim_{\|P\|\to 0}\sum_{i=1}^n f(x_i^*)\,\Delta x_i.

Because every sum in the limit is nonnegative, the limit itself must be nonnegative.

Therefore, ∫abf(x)dx≥0.\int_a^b f(x)\,dx \ge 0.

∫abf(x)dx≥0\boxed{\int_a^b f(x)\,dx \ge 0}

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