Use definite integrals and Fundamental Theorem of Calculus to find the exact area of the region between the graphs of $f ( x ) = x ^ { 2 }$ and $g ( x ) = x + 6$
A) 21
B) 123/6
C) 125/6
D) 124/6

Question

Quizplus · Accepted Answer

To find the area between the graphs of $f(x) = x^2$ and $g(x) = x + 6$, we first need to find the points of intersection by setting $f(x) = g(x)$, which gives us $x^2 = x + 6$. Solving this quadratic equation, we get $x^2 - x - 6 = 0$, which factors to $(x-3)(x+2) = 0$, giving us the points of intersection at $x = 3$ and $x = -2$. The area between the curves from $x = -2$ to $x = 3$ is given by the integral of the upper function minus the lower function over this interval. Thus, the area $A$ is given by:$$A = \int_{-2}^{3} (g(x) - f(x)) \, dx = \int_{-2}^{3} ((x + 6) - x^2) \, dx$$$$A = \int_{-2}^{3} (-x^2 + x + 6) \, dx$$$$A = \left[ -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 6x \right]_{-2}^{3}$$$$A = \left( -\frac{1}{3}(3)^3 + \frac{1}{2}(3)^2 + 6(3) \right) - \left( -\frac{1}{3}(-2)^3 + \frac{1}{2}(-2)^2 + 6(-2) \right)$$$$A = \left( -9 + \frac{9}{2} + 18 \right) - \left( \frac{8}{3} + 2 - 12 \right)$$$$A = \left( -9 + 4.5 + 18 \right) - \left( \frac{8}{3} + 2 - 12 \right)$$$$A = 13.5 - \left( \frac{8}{3} - 10 \right)$$$$A = 13.5 - \left( \frac{8}{3} - \frac{30}{3} \right)$$$$A = 13.5 + \frac{22}{3}$$$$A = \frac{81}{6} + \frac{44}{6}$$$$A = \frac{125}{6}$$Therefore, the exact area of the region between the graphs of $f(x) = x^2$ and $g(x) = x + 6$ is $\frac{125}{6}$.

Use Definite Integrals and Fundamental Theorem of Calculus to Find f(x)=x2f ( x ) = x ^ { 2 }f(x)=x2

Use Definite Integrals and Fundamental Theorem of Calculus to Find $f ( x ) = x ^ { 2 }$