Observe the roots, don't expand

The solution to this problem is rather simple if you were to use a certain property of definite integrals:

$\int_\alpha^\beta (x -\alpha)(x - \beta) dx = - \frac{1}{6}(\beta-\alpha)^{3}$

Here's the proof:

$\int_\alpha^\beta (x - \alpha)(x- \beta) dx$

$= \int_\alpha^\beta [x^{2} - (\alpha + \beta)x + \alpha\beta] dx$

$= [\frac{1}{3}x^{3}]_\alpha^\beta - (\alpha + \beta)[\frac{1}{2}x^{2}]_\alpha^\beta + \alpha\beta[x]_\alpha^\beta$

$= \frac{1}{3}(\beta^{3} - \alpha^{3}) - \frac{1}{2}(\alpha + \beta)(\beta^{2} - \alpha^{2}) + \alpha\beta(\beta - \alpha)$

$= \frac{1}{6}(\beta - \alpha)[2(\beta^{2} + \alpha\beta + \alpha^{2}) - 3(\beta + \alpha)^{2} + 6\alpha\beta]$

$= \frac{1}{6}(\beta - \alpha)(-\beta^{2} + 2\alpha\beta - \alpha^{2})$

$= -\frac{1}{6}(\beta - \alpha)^{3}$

As such, we have

$\int_1^7 (x^{2} - 8x + 7) dx = - \frac {1}{6}(7-1)^{3} = \boxed{-36}$

Slightly overengineered solution: make the substitution

$x = y+4 \\ dx = dy \\ \int^3_{-3}{(y+3)(y-3)} dy \\ 2\int^3_0{(y+3)(y-3)} dy \\ 2\int^3_0{y^2-9} dy \\ 2(\frac{y^3}{3}-9y)|^3_0 \\ 2(9-27) \\ 2(-18) = \boxed{-36}$

Makes it a bit easier to do mentally.

I did it the slow way, expand, $\int_{1}^{7}(x^{2}-8x+7)dx$ , then use reverse power rule, to integrate, which produces $\frac{1}{3}\cdot x^{3}-4\cdot x^{2}+x\cdot 7$ . After that, substitute in 7 into the expression, and subtract the result of substituting 1 into the expression. $\left (\frac{1}{3}\cdot 7^{3}-4\cdot 7^{2}+7\cdot 7 \right )-\left (\frac{1}{3}\cdot 1^{3}-4\cdot 1^{2}+7\cdot 1 \right )$ This produces the result $-36$