Find the value of x. 2(x-3) x 8(x+2) = 128

$$\int 2x \, dx = x^2 + C$$ $$\left[ x^2 \right]_1^5 = 5^2 - 1^2$$ $$5^2 = 25, \quad 1^2 = 1$$ $$25 - 1 = 24$$

$\int_1^5 2x \, dx = 24$

24

$$\int (2x + 3) \, dx = x^2 + 3x + C$$ $$\left[ x^2 + 3x \right]_1^5 = (5^2 + 3 \cdot 5) - (1^2 + 3 \cdot 1)$$ $$5^2 + 3 \cdot 5 = 25 + 15 = 40$$ $$1^2 + 3 \cdot 1 = 1 + 3 = 4$$ $$40 - 4 = 36$$

$\int_1^5 (2x + 3) \, dx = 36$

36

$$\int (x^2 - 3x + 2) \, dx = \frac{1}{3} x^3 - \frac{3}{2} x^2 + 2x + C$$ $$\left[ \frac{1}{3} x^3 - \frac{3}{2} x^2 + 2x \right]_1^5$$

At $x=5$:

$$\frac{1}{3} (125) - \frac{3}{2} (25) + 2(5) = \frac{125}{3} - \frac{75}{2} + 10$$

Common denominator 6:

$$\frac{125}{3} = \frac{250}{6}, \quad \frac{75}{2} = \frac{225}{6}, \quad 10 = \frac{60}{6}$$ $$\frac{250}{6} - \frac{225}{6} + \frac{60}{6} = \frac{250 - 225 + 60}{6} = \frac{85}{6}$$

At $x=1$:

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

Common denominator 6:

$$\frac{1}{3} = \frac{2}{6}, \quad \frac{3}{2} = \frac{9}{6}, \quad 2 = \frac{12}{6}$$ $$\frac{2}{6} - \frac{9}{6} + \frac{12}{6} = \frac{2 - 9 + 12}{6} = \frac{5}{6}$$ $$\frac{85}{6} - \frac{5}{6} = \frac{80}{6} = \frac{40}{3}$$

$\int_1^5 (x^2 - 3x + 2) \, dx = \dfrac{40}{3}$

\dfrac{40}{3}