Question

Verify the second mean value theorem of integrability for the functions f and g
defined on [1,2] by f (x) = 3x and g(x) = 5x.

Answer 1

Answer on Question #64862 – Math – Real Analysis

Question

Verify the second mean value theorem of integrability for the functions $f$ and $g$ defined on $[1, 2]$ by $f(x) = 3x$ and $g(x) = 5x$.

Solution

Let us state the second mean value theorem.

If $f \colon [a, b] \to \mathbb{R}$ is a monotonic function and $g \colon [a, b] \to \mathbb{R}$ is an integrable function, then there exists a number $x$ in $(a, b)$ such that

$$\int_{a}^{b} f(t) g(t) dt = f(a^{+}) \int_{a}^{x} g(t) dt + f(b^{-}) \int_{x}^{b} g(t) dt.$$

Function $f(x) = 3x$ increases on $[1,2]$;

$$\int_{1}^{2} g(t) dt = \int_{1}^{2} 5t dt = \frac{5t^{2}}{2} \bigg|_{1}^{2} = \frac{15}{2} < \infty, \text{ so } g(x) = 5x \text{ is integrable on } [1, 2].$$

Then

$$\begin{aligned} \int_{1}^{2} 15t^{2} dt &= f(1^{+}) \int_{1}^{x} 5t dt + f(2^{-}) \int_{x}^{2} 5t dt; \\ \int_{1}^{2} 15t^{2} dt &= 15 \int_{1}^{x} t dt + 30 \int_{x}^{2} t dt; \\ 5t^{3} \bigg|_{t=1}^{2} &= \frac{15t^{2}}{2} \bigg|_{t=1}^{x} + 15t^{2} \bigg|_{t=x}^{2}; \\ 35 &= \frac{15x^{2}}{2} - \frac{15}{2} + 60 - 15x^{2}. \end{aligned}$$

We must solve this equation on $(1,2)$.

$$- \frac{15x^{2}}{2} = - \frac{35}{2} \Rightarrow x^{2} = \frac{7}{3} \Rightarrow x = \sqrt{\frac{7}{3}} \in (1, 2).$$

So we found $x = \sqrt{\frac{7}{3}} \in (1, 2)$ such that

$$\int_{1}^{2} f(t) g(t) dt = f(1^{+}) \int_{1}^{x} g(t) dt + f(2^{-}) \int_{x}^{2} g(t) dt.$$

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