# Thread: Integral and Graph Quick Help

1. ## Integral and Graph Quick Help

In the pic above, why is there a +8 for problem number 3? I thought it would be just 5x-8. How do you calculate the 8?
I know A'(x) = f(x), and that A(x) = integral of f(x)

2. ## Re: Integral and Graph Quick Help

There are many integrals for f(x). Infinitely many, in fact. You need to use an initial condition to determine the correct constant of integration. You have:

$\displaystyle f(x) = \begin{cases}3 & 0\le x \le 3 \\ x & 3 \le x \le 5 \\ 5 & 5 \le x \le 7\end{cases}$

So, the integral of f on [5,7] is:

$\displaystyle A(x) = 5x+C$

On the interval [3,5], you have:

$\displaystyle A(5) = \dfrac{5^2}{2}+\dfrac{9}{2} = 17$

So, now, you want $\displaystyle A(5) = 5(5)+C = 17 \Longrightarrow C = -8$.

Basically, at each point between two intervals, you want the A(x) function to be continuous, so you find the value of the constant of integration to make it continuous.

3. ## Re: Integral and Graph Quick Help

Originally Posted by SlipEternal
There are many integrals for f(x). Infinitely many, in fact. You need to use an initial condition to determine the correct constant of integration. You have:

$\displaystyle f(x) = \begin{cases}3 & 0\le x \le 3 \\ x & 3 \le x \le 5 \\ 5 & 5 \le x \le 7\end{cases}$

So, the integral of f on [5,7] is:

$\displaystyle A(x) = 5x+C$

On the interval [3,5], you have:

$\displaystyle A(5) = \dfrac{5^2}{2}+\dfrac{9}{2} = 17$

So, now, you want $\displaystyle A(5) = 5(5)+C = 17 \Longrightarrow C = -8$.

Basically, at each point between two intervals, you want the A(x) function to be continuous, so you find the value of the constant of integration to make it continuous.
Alright, thanks! Didn't realize that A(x) should be continous and that C is a value we must define!!