find the area, in square units, bounded above by f(x)=x^2 + 2x + 1 and below by g(x)=2x + 10 over the…

Explanation:

Step1: Recall area - between - curves formula

The area (A) between two curves (y = f(x)) and (y = g(x)) over the interval ([a,b]) is given by (A=\int_{a}^{b}|f(x)-g(x)|dx). Here, (f(x)=x^{2}+2x + 1), (g(x)=2x + 10), (a=-5), and (b = - 1). First, find (f(x)-g(x)): [ \begin{align*} f(x)-g(x)&=(x^{2}+2x + 1)-(2x + 10)\ &=x^{2}+2x + 1-2x - 10\ &=x^{2}-9 \end{align*} ]

Step2: Set up the integral

We need to calculate (\int_{-5}^{-1}|x^{2}-9|dx). To deal with the absolute - value, find where (y=x^{2}-9=(x + 3)(x - 3)) is non - negative and non - positive on the interval ([-5,-1]). Set (x^{2}-9 = 0), then (x=-3) or (x = 3). On the interval ([-5,-3]), (x^{2}-9\geq0), and on the interval ([-3,-1]), (x^{2}-9\leq0). So, (\int_{-5}^{-1}|x^{2}-9|dx=\int_{-5}^{-3}(x^{2}-9)dx+\int_{-3}^{-1}-(x^{2}-9)dx).

Step3: Integrate term - by - term

The antiderivative of (x^{2}-9) is (\frac{1}{3}x^{3}-9x). For (\int_{-5}^{-3}(x^{2}-9)dx=\left[\frac{1}{3}x^{3}-9x\right]{-5}^{-3}) [ \begin{align*} &=\left(\frac{1}{3}(-3)^{3}-9(-3)\right)-\left(\frac{1}{3}(-5)^{3}-9(-5)\right)\ &=\left(-9 + 27\right)-\left(-\frac{125}{3}+45\right)\ &=18-\left(-\frac{125}{3}+\frac{135}{3}\right)\ &=18-\frac{10}{3}\ &=\frac{54 - 10}{3}=\frac{44}{3} \end{align*} ] The antiderivative of (-(x^{2}-9)=-x^{2}+9) is (-\frac{1}{3}x^{3}+9x). For (\int{-3}^{-1}-(x^{2}-9)dx=\left[-\frac{1}{3}x^{3}+9x\right]_{-3}^{-1}) [ \begin{align*} &=\left(-\frac{1}{3}(-1)^{3}+9(-1)\right)-\left(-\frac{1}{3}(-3)^{3}+9(-3)\right)\ &=\left(\frac{1}{3}-9\right)-\left(9 - 27\right)\ &=\frac{1 - 27}{3}-(-18)\ &=-\frac{26}{3}+18\ &=\frac{-26 + 54}{3}=\frac{28}{3} \end{align*} ]

Step4: Sum the results of the two integrals

(A=\frac{44}{3}+\frac{28}{3}=\frac{44 + 28}{3}=\frac{72}{3}=24)

Answer:

24