To find the area of the shaded region bounded by the functions:

$f(x) = -x^2 + 1.75x + 10$

$g(x) = 1.5x + 7$

Step 1: Find the Points of Intersection

We set $f(x) = g(x)$ to determine the points where the curves intersect:

$- x^2 + 1.75x + 10 = 1.5x + 7$

Rearrange the equation:

$- x^2 + 1.75x + 10 - 1.5x - 7 = 0$

$- x^2 + 0.25x + 3 = 0$

Solve for $x$ using the quadratic formula:

$x = \frac{-0.25 \pm \sqrt{(0.25)^2 - 4(-1)(3)}}{2(-1)}$

$x = \frac{-0.25 \pm \sqrt{0.0625 + 12}}{-2}$

$x = \frac{-0.25 \pm \sqrt{12.0625}}{-2}$

Approximating $\sqrt{12.0625} \approx 3.47$:

$x = \frac{-0.25 \pm 3.47}{-2}$

Finding the two values of $x$:

$x_1 = \frac{-0.25 + 3.47}{-2} = \frac{3.22}{-2} \approx -1.61$

$x_2 = \frac{-0.25 - 3.47}{-2} = \frac{-3.72}{-2} \approx 1.86$

Step 2: Compute the Area

The area of the region is given by:

$A = \int_{x_1}^{x_2} [f(x) - g(x)] \,dx$

$A = \int_{-1.61}^{1.86} [(-x^2 + 1.75x + 10) - (1.5x + 7)] \,dx$

$A = \int_{-1.61}^{1.86} (-x^2 + 0.25x + 3) \,dx$

Computing the integral:

$\int (-x^2 + 0.25x + 3) \,dx = -\frac{x^3}{3} + \frac{0.25x^2}{2} + 3x$

Evaluating at $x_2 = 1.86$:

$-\frac{(1.86)^3}{3} + \frac{0.25(1.86)^2}{2} + 3(1.86)$

$= -\frac{6.44}{3} + \frac{0.25(3.46)}{2} + 5.58$

$= -2.15 + 0.43 + 5.58 = 3.86$

Evaluating at $x_1 = -1.61$:

$-\frac{(-1.61)^3}{3} + \frac{0.25(-1.61)^2}{2} + 3(-1.61)$

$= -\frac{-4.17}{3} + \frac{0.25(2.59)}{2} - 4.83$

$= 1.39 + 0.32 - 4.83 = -3.12$

Now, subtract:

$A = 3.86 - (-3.12) = 3.86 + 3.12 = 6.98$

Final Answer:

$\boxed{6.980}$

Would you like more details or have any questions?

Related Questions:

1. How do you set up definite integrals for area between curves?
2. What is the significance of the intersection points in area calculations?
3. How do you approximate square roots manually?
4. How does integration differ from summation in finding areas?
5. Can we use a calculator to directly compute this definite integral?

Tip:

Always double-check your integration limits and expressions before solving!