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Answer:
[tex]\displaystyle \int^{6}_{-4} \left(-x^2+2x+24\right)\; dx[/tex]
Step-by-step explanation:
To find the area of the region between the graphs of y = x² - 11 and y = 2x + 13, we need to determine the points of intersection and then set up the definite integral for the area between these curves.
To find the points of intersection, set the equations equal to each other and solve for x:
[tex]x^2-11=2x+13 \\\\x^2-2x-24=0 \\\\ x^2-6x+4x-24=0 \\\\ x(x-6)+4(x-6)=0 \\\\ (x+4)(x-6)=0\\\\\\x+4=0 \implies x=-4 \\\\x-6=0 \implies x=6[/tex]
Therefore, the values of x where the curves intersect are x = -4 and x = 6.
The area R between the curves on the interval [a, b] is given by the integral of the difference between the upper function and the lower function:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Area between two curves}}\\\\\displaystyle A = \int_{a}^{b} [f(x) - g(x)] \; dx\\\\\textsf{where:}\\\phantom{ww} \bullet \; \textsf{$y=f(x)$ is the upper function.} \\ \phantom{ww} \bullet \; \textsf{$y=g(x)$ is the lower function.} \\ \phantom{ww} \bullet \; \textsf{$[a, b]$ is the interval.}\end{array}}[/tex]
In this case, the linear function is the upper function and the quadratic function is the lower function between the points of intersection:
[tex]f(x) = 2x + 13[/tex]
[tex]g(x) = x^2 - 11[/tex]
[tex]a = -4[/tex]
[tex]b = 6[/tex]
Therefore:
[tex]\displaystyle R=\int^{6}_{-4} \left((2x+13)-(x^2-11)\right)\; dx[/tex]
Simplify the integrand:
[tex]\displaystyle R=\int^{6}_{-4} \left(2x+13-x^2+11\right)\; dx \\\\\\ R=\int^{6}_{-4} \left(-x^2+2x+24\right)\; dx[/tex]
Therefore, the definite integral to find the area of the region between the graphs of y = x² - 11 and y = 2x + 13 is:
[tex]\Large\boxed{\boxed{\int^{6}_{-4} \left(-x^2+2x+24\right)\; dx}}[/tex]
