Integral (1) fixed

$\begin{aligned} I & = \int_7^{17} \left(x^3-29x^2+281x-890 \right) dx \\ & = \int_7^{17} \left(x^3-21x^2+147x+343-8x^2+134x-547 \right) dx \\ & = \int_7^{17} \left((x-7)^3-8(x^2-14x+49) + 22x-154-1 \right) dx \\ & = \int_7^{17} \left((x-7)^3-8(x-7)^2+22(x-7)-1 \right) dx & \small \color{#3D99F6} \text{Let }u = x-7 \\ & = \int_0^{10} \left(u^3-8u^2+22u-1 \right) du \\ & = \frac {u^4}4 - \frac {8u^3}3 + 11u - u \bigg|_0^{10} \\ & = \frac {2770}3 \approx \boxed{923.333} \end{aligned}$

$x^3-29x^2+281x-890=(x-10)^3+(x-10)^2+(x-10)$ \ Now sub $u=x-10$ and we integrate from -7 to 7 So cubic and linear vanish becomes $\frac{(x-10)^3}{3}$ from -7 to 7 $\boxed{\frac{686}{3}}$

This was if lower bound was 3...