# Find $\int_1^3f(x)\ dx$ if $\int_0^1f(x)\ dx$ , $\int_0^4f(x)\ dx$ and $\int_3^4f(x)\ dx$ are given

Say I have some integrals:

$$\int_0^1f(x)\ dx=3$$ $$\int_0^4f(x)\ dx=-5$$ $$\int_3^4f(x)\ dx=1$$

How exactly do I go about evaluating the integral for:

$$\int_1^3f(x)\ dx$$

My idea is I must take the area of each individual segment and add them together. Or average. I am lost here.

• You're pretty much correct; what did you try already? Remember also that you can subtract areas using the rule $\int_a^c f - \int_b^c f = \int_a^b f$. Commented Jun 5, 2017 at 4:30
• I tried adding all the values and getting -1, then dividing it by 3 segments. I understand though each integral is involving different segments and after drawing a rough sketch of some bars from 0 to 4, I find I have no value for the point 1 to 2 Commented Jun 5, 2017 at 4:32
• Don't worry about division - try adding and subtracting to get cancellation. You won't need a sketch in this case (although drawing one is a good thing to try in general)! Commented Jun 5, 2017 at 4:33
• Ah -9, wow thanks Commented Jun 5, 2017 at 4:46
• Commented Jun 5, 2017 at 8:11

## 4 Answers

Sketch a diagram and convince yourself this must indeed be the case:

$$\int_0^4 = \int_0^1 + \int_1^3 + \int_3^4$$

Hint: Take the area of the entire interval ($0$ to $4$), then subtract from it the two extra areas that you don't want.

• Very helpful, thank you! Commented Jun 5, 2017 at 4:46

Use the fact that $$\int_{a}^{b} f(x)dx + \int_{b}^{c} f(x)dx = \int_{a}^{c} f(x)dx$$ In this case, $$\int_{0}^{1} f(x)dx + \int_{1}^{3} f(x)dx + \int_{3}^{4} f(x)dx = \int_{0}^{4} f(x)dx$$ This becomes a simple matter of solving for $\int_{1}^{3} f(x)dx$, $$\int_{1}^{3} f(x)dx = \int_{0}^{4} f(x)dx - \int_{0}^{1} f(x)dx - \int_{3}^{4} f(x)dx$$ which yields the desired answer.

A longer path can be developed by considering a form for $f(x)$. Since there are three equations then consider a three coefficient polynomial, ie $f(x) = a + b \, x + c \, x^2$. Upon integration it is seen that: \begin{align} 3 &= a + \frac{b}{2} + \frac{c}{3} \\ -5 &= 4 \, a + 8 \, b + \frac{64 \, c}{3} \\ 1 &= a + \frac{7 \, b}{2} + \frac{37 \, c}{3} \end{align} which, when solved, the function $f(x)$ takes the form $$f(x) = \frac{1}{12} \, (105 - 164 \, x + 39 \, x^2).$$

Now, \begin{align} \int_{1}^{3} f(x) \, dx = \frac{1}{12} \, \left[ 105 \, x - 82 \, x + 13 \, x^2 \right]_{1}^{3} = -9. \end{align}

In comparison: $$\int_{1}^{3} f(x) \, dx = \left( - \int_{0}^{1} + \int_{0}^{4} - \int_{3}^{4} \right) \, f(x) \, dx = -3 -5 -1 = -9.$$