<s>[INST] Given several integrals calculate $\int\limits_5^6 f(x) dx$ Let
$\int\limits_4^7 f(x)\,dx = 2$,  $\int\limits_6^7 f(x)\,dx = 17$, and  $\int\limits_4^5 f(x)\,dx = 3$
Calculate
$$\int\limits_5^6 f(x)\,dx$$
I guess I am to assume that $f(x)$ is the same in all integrals.
Should I figure out what $\Delta x$ is? Should I try to use identities of integrals?
 [/INST] Note: Although this problem is older the method displayed is useful as a secondary demonstration.
The problem presents three equations and asks to evaluate a fourth. Since the results are simple the form of the integrand can be considered in a simple form, ie $f(x) = a + b \, x + c \, x^2$. In this view it can be seen that:
$$\int f(x) \, dx = a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3}$$
for which
\begin{align}
\int_{4}^{7} f(x) \, dx = 2 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3} \\
\int_{6}^{7} f(x) \, dx = 17 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3} \\
\int_{4}^{5} f(x) \, dx = 3 &= a \, x + \frac{b \, x^2}{2} + \frac{c \, x^3}{3}. 
\end{align}
Solving this system f equations yields
$$f(x) = \frac{1}{6} \, ( 4729 - 1806 \, x + 168 \, x^2)$$
and leads to
$$\int_{5}^{6} f(x) \, dx = \frac{1}{6} \, [ 4729 \, x - 1403 \, x^2 + 56 \, x^3]_{5}^{6} = -18.$$
In comparison it is seen that
$$\int_{5}^{6} f(x) \, dx = \left(\int_{4}^{7} - \int_{6}^{7} - \int_{4}^{5} \right) \, f(x) \, dx = (2 - 17 - 3) = -18.$$
An advantage for finding and solving the system of equations is by considering if the proposed question had asked for the value of the integral for the range (0,10), or (0,a), etc..
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