Set up an integral or sum of integrals and then find the area bounded by y = $x ^ { 2 }$ + 3 and y = -5x - 3.
A) $\int _ { - 3 } ^ { - 2 } \left[ \left( x ^ { 2 } + 3 \right) - ( - 5 x - 3 ) \right] d x$ = - $\frac { 1 } { 6 }$
B) $\int _ { 2 } ^ { 3 } \left[ \left( x ^ { 2 } + 3 \right) - ( - 5 x - 3 ) \right] d x$ = $\frac { 149 } { 6 }$
C) $\int _ { - 3 } ^ { - 2 } \left[ ( - 5 x - 3 ) - \left( x ^ { 2 } + 3 \right) \right] d x$ = $\frac { 1 } { 6 }$
D) $\int _ { 2 } ^ { 3 } \left[ ( - 5 x - 3 ) - \left( x ^ { 2 } + 3 \right) \right] d x$ = - $\frac { 149 } { 6 }$
E) none of these

Question

Quizplus · Accepted Answer

The correct integral setup is to find the area between the curves by integrating the difference of the functions from the left intersection point to the right intersection point. The integral in option C correctly represents the area between the curves from x = -3 to x = -2.

Set Up an Integral or Sum of Integrals and Then x2x ^ { 2 }x2

Set Up an Integral or Sum of Integrals and Then $x ^ { 2 }$