for each problem, find the area of the region enclosed by the curves. you may use the provided graph to…

for each problem, find the area of the region enclosed by the curves. you may use the provided graph to sketch the curves and shade the enclosed region.\n\n5) $y = -2x^2 - 1$\n$y = -x + 3$\n$x = 0$\n$x = 1$

for each problem, find the area of the region enclosed by the curves. you may use the provided graph to sketch the curves and shade the enclosed region.\n\n5) $y = -2x^2 - 1$\n$y = -x + 3$\n$x = 0$\n$x = 1$

Answer

Explanation:

Step1: Identify the upper and lower functions

On the interval $[0, 1]$, $y = -x + 3$ is greater than $y = -2x^2 - 1$. $$f(x) = -x + 3, \quad g(x) = -2x^2 - 1$$

Step2: Set up the definite integral for area

The area $A$ is the integral of the upper minus lower function. $$A = \int_{0}^{1} [(-x + 3) - (-2x^2 - 1)] , dx$$

Step3: Simplify the integrand

Combine like terms within the integral. $$A = \int_{0}^{1} (2x^2 - x + 4) , dx$$

Step4: Find the antiderivative

Apply the power rule for integration. $$\int (2x^2 - x + 4) , dx = \frac{2}{3}x^3 - \frac{1}{2}x^2 + 4x$$

Step5: Evaluate at the boundaries

Substitute the limits $x=1$ and $x=0$ into the antiderivative. $$A = \left[ \frac{2}{3}(1)^3 - \frac{1}{2}(1)^2 + 4(1) \right] - [0]$$

Step6: Calculate the final value

Find a common denominator to sum the fractions. $$A = \frac{2}{3} - \frac{1}{2} + 4 = \frac{4 - 3 + 24}{6} = \frac{25}{6}$$

Answer:

$\frac{25}{6}$