Integral Computation check

Question

Question

Let $f:\Bbb{R}\to\Bbb{R}$ differentiable and $f(2)=3$ and $1 \geq f'(x)$ for every $x$.Find the best estimation of $$ \int_{2}^{5} f(x)dx $$

Solution

$$\frac{f(5)-f(2)}{3}\leq 1\tag{Intermediate Value Theorem}$$
Now I change $u=f(x)$ , $du=f'(x)dx$ so $$\int_{f(2)}^{f(5)}f'(x)udu \leq\int_{f(2)}^{f(5)}udu \leq\int_{3}^{6}udu=\frac{25}{2} $$ Can I find a better estimation?

Answer 1

$$f'(x)\le 1\implies f(x)\le x+C$$ and $$f(2)\le2+C=3.$$

So if the bound is tight,

$$\int_2^5(x+1)\,dx=\left.\frac{(x+1)^2}2\right|_2^5=\frac{27}2.$$

I wouldn't call this the "best" estimation. It is rather the only one we can make.

Answer 2

Your final number should be $27/2$, and this is the best you can do. A more elementary way to get the number is to draw the trapezoid with vertices at $(2,0), (2,3), (5,0),$ and $(5,6)$. The top edge of the trapezoid is an upper bound for $f(x)$, since it's a line of slope $1$. The area of the trapezoid is $27/2$ and the integral must be less than that. Since the top edge satisfies the conditions for $f(X)$, you know you can't do better.