Question

Find the area of the region bounded by the graphs of y = −x^2 + 3x + 4 and y = 4.

Answer 1

a. 2.7
b. 4.5
c. 28.5
d. 16.5

Answer 2

@rational

Answer 3

the area under the curve of the entire parabola is
\(\large\color{slate}{\displaystyle\int\limits_{0}^{3}\left(-x^2+3x+4\right)~dx}\)

and subtract the \(\large\color{slate}{\displaystyle\int\limits_{0}^{3}\left(4\right)~dx}\) from that

Answer 4

Or, you can just shift the parabola 4 units down, and your area is

\(\large\color{slate}{\displaystyle\int\limits_{0}^{3}\left(-x^2+3x\right)~dx}\)

Answer 5

4.5 b

Answer 6

@SolomonZelman can u help me on one more ?

Answer 7

yes, 4.5 is correct

Answer 8

I can try, I suck at math

Answer 9

Find the area or the region bounded by the curves y = x^3 and y = 9x.
a. 0
b. 10.13
c. 40.50
d. 20.25

Answer 10

subtract the areas.

first we need to solve for the second intersection point (besides x=0)

Answer 11

(0,0) and (3,27) is the intersection
(found it on the graph)

where 9x >= x^3 on out interval of [0,3]
so you can just subtract the areas

Answer 12

\(\large\color{slate}{\displaystyle\int\limits_{0}^{3}\left(9x\right)~dx-\int\limits_{0}^{3}\left(x^3\right)~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{0}^{3}\left(9x- x^3\right)~dx}\)

Answer 13

on our (not out) interval ***

Answer 14

20.25

Answer 15

yes, correct