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.
 A: Sketch a diagram and convince yourself this must indeed be the case:
$$\int_0^4 = \int_0^1 + \int_1^3 + \int_3^4$$
A: Hint: Take the area of the entire interval ($0$ to $4$), then subtract from it the two extra areas that you don't want.
A: 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: 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:
$$ \int_{1}^{3} f(x) \, dx = \left( - \int_{0}^{1} + \int_{0}^{4} - \int_{3}^{4} \right) \, f(x) \, dx = -3 -5 -1 = -9. $$
