Calculus find extreme values of integral

Question

I have this problem:

Ineach of Exercises 48-51, a definite integral is given. Do not attempt to calculate its value $V$. Instead, find the extreme values of the integrand on the interval of integration, and use these extreme values together with the inequalities of line (5.3.5) to obtain numbers $A$ and $B$ such that $A \le V \le B$.

$\displaystyle \int_{-1}^2 \frac{x^2+5}{x+2} \, dx$

$\displaystyle m\cdot(b-a) \le \int_a^b f(x) \, dx \le M \cdot (b-a)$

I'm not 100% as what to do. It says don't solve, so what do I do?

Answer 1

On the interval $[-1,2]$, perhaps you can find the minimum of $f(x)=\frac{x^2+5}{x+2}$. If you evaluate $$\frac d{dx}\frac{x^2+5}{x+2}$$ and set it equal to $0$, you would find the vertex to occur at $x=1$, and the corresponding $y$-value to be $2$. So the minimum value, $m$, of the function $\frac{x^2+5}{x+2}$ on the interval $[-1,2]$ is $m=f(1)=2$. Thus, we have $$A=m\cdot(b-a)=2\cdot(2--1)=6.$$

Now on the same interval $[-1,2]$, the maximum occurs at the left endpoint of the interval, $x=-1$. That is, $f(-1)=6$. (And the right endpoint, $f(2)=2.25$, which is not the maximum of $f(x)$ on the interval $[-1,2]$.) So $M=f(-1)=6$. Thus, $$B=M \cdot (b-a) = 6\cdot(2--1)=18.$$

Here is a visual representation of what I mean:

Answer 2

$$f(x)=\frac{x^2+5}{x+2}\implies f'(x)=\frac{(x+5)(x-1)}{(x+2)^2}$$

We can see the function has a minimal value at $\;x=1\;:\;\;f(1)=2\;$, and since it is decreasing in $\;(-1,1)\;$ and increasing in $\;(1,2)\;$ , it's maximum is either $\;f(-1)=6\;$ or $\;f(2)=\frac94\;$ .

Thus, $\;M=6\;,\;\;m=2\;$ and now you can find $\;A,B\;$ as asked.

Answer 3

Here $m$ represents the minimal value that $\frac{x^2+5}{x+2}$ takes on the interval $[-1,2]$ and $M$ represents the maximum. The problem boils down to finding $m$ and $M$. The $(b-a)$ that appears on the left and right sides of the inequality is simply the length of the interval you are integrating over. The purpose of multiplying that length times $m, M$ respectively is to create rectangles that act as lower and upper bounds of the actual area of your integral. So let's find $m$ and $M$. we can do this by setting the derivative of the function equal to zero and finding the local min/max that occurs on $[-1,2]$. It might help to note that $\frac{x^2+5}{x+2} = x-2+\frac{9}{x+2}$. Then $$\left(x-2+\frac{9}{x+2}\right)'=1-\frac{9}{(x+2)^2}$$ and $$1-\frac{9}{(x+2)^2}=0 \\ \implies 1=\frac{9}{(x+2)^2} \\ \implies (x+2)^2=9 \\ \implies x+2 = \pm 3 \\ \implies x = 1,-5$$ But $-5$ is outside our interval so we only care about $x=1$. It should be easy to verify that a local minimum occurs at $x=1$. Thus, we know that $f(1) = 2$ is our minimum, so $m=2$. Now we find $M$. We couldn't find $M$ using the first derivative, so $M$ must occur at one of the boundaries. A quick plug and check reveals that $f(-1) = 6$ and $f(2) = \frac{9}{4}$ so $M=6$. Now that we have our min and max on $[-1,2]$, we may conclude $$2(2-(-1)) \leq \int_{-1}^2 \frac{x^2+5}{x+2} dx \leq 6(2-(-1)) \\ \implies 6 \leq \int_{-1}^2 \frac{x^2+5}{x+2} dx \leq 18$$ If you were to actually calculate the integral, you would find $$\int_{-1}^2 \frac{x^2+5}{x+2} dx = \frac{x^2}{2}-2x+9\ln\left|x+2\right|\Bigg |_{-1}^2 \\ =(2-2+9\ln|4|)-(\frac{1}{2}+2+9\ln|1|) \\ = 9\ln(4)-\frac{5}{2} \\ \approx 9.98$$ We have verified that we found the right $m,M$ as it is indeed true that $$6 \leq 9.98 \leq 18$$