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Comparing integrals using the dominance property

Let f(x)=exf(x) = e^x and g(x)=x+1g(x) = x+1 on the interval [0,2][0,2].

First note that for all xx, ex≥x+1e^x \ge x+1 (the exponential function lies above its tangent line at x=0x=0), so f(x)≥g(x)f(x) \ge g(x) for every xx in [0,2][0,2].

Using the dominance property of definite integrals (without computing either integral explicitly), what can you conclude about the relationship between ∫02ex dx\displaystyle\int_0^2 e^x\,dx and ∫02(x+1) dx\displaystyle\int_0^2 (x+1)\,dx? Write your conclusion as an inequality involving the two integrals.

Hint

Recall the dominance property: if f(x)≥g(x)f(x) \ge g(x) for all xx in [a,b][a,b], then ∫abf(x) dx≥∫abg(x) dx\int_a^b f(x)\,dx \ge \int_a^b g(x)\,dx. Apply this directly to ff and gg here.

Step-by-step solution

Since f(x)=exf(x) = e^x and g(x)=x+1g(x) = x+1 satisfy f(x)≥g(x)f(x) \ge g(x) for all x∈[0,2]x \in [0,2], the dominance property of definite integrals tells us that the integral of the larger function over the same interval must be at least as large as the integral of the smaller function.

Applying this property directly with a=0a=0 and b=2b=2 gives ∫02ex dx≥∫02(x+1) dx\int_0^2 e^x\,dx \ge \int_0^2 (x+1)\,dx. This conclusion follows purely from the pointwise inequality f≥gf \ge g, with no need to evaluate either integral.

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