question. determine the area, in square units, bounded above by f(x)=x² + 10x + 25 and g(x)=-x + 1 and…

Explanation:

Step1: Determine the upper - lower functions

Since (f(x)=x^{2}+10x + 25=(x + 5)^{2}\geq0) and (g(x)=-x + 1) on the interval ([-5,1]), the upper function is (y_1=x^{2}+10x + 25) and the lower function is (y_2=-x + 1).

Step2: Use the area formula

The area (A) between two curves (y = y_1) and (y = y_2) on the interval ([a,b]) is given by (A=\int_{a}^{b}(y_1 - y_2)dx). Here, (a=-5), (b = 1), (y_1=x^{2}+10x + 25), and (y_2=-x + 1). So (A=\int_{-5}^{1}[(x^{2}+10x + 25)-(-x + 1)]dx=\int_{-5}^{1}(x^{2}+11x + 24)dx).

Step3: Integrate term - by - term

We know that (\int(x^{2}+11x + 24)dx=\frac{1}{3}x^{3}+\frac{11}{2}x^{2}+24x+C).

Step4: Evaluate the definite integral

[ \begin{align*} A&=\left[\frac{1}{3}x^{3}+\frac{11}{2}x^{2}+24x\right]_{-5}^{1}\ &=\left(\frac{1}{3}(1)^{3}+\frac{11}{2}(1)^{2}+24(1)\right)-\left(\frac{1}{3}(-5)^{3}+\frac{11}{2}(-5)^{2}+24(-5)\right)\ &=\left(\frac{1}{3}+\frac{11}{2}+24\right)-\left(-\frac{125}{3}+\frac{275}{2}-120\right)\ &=\frac{2 + 33+144}{6}-\left(\frac{-250 + 825-720}{6}\right)\ &=\frac{179}{6}-\left(\frac{-145}{6}\right)\ &=\frac{179 + 145}{6}\ &=\frac{324}{6}\ & = 54 \end{align*} ]

Answer:

54