Question

@stamp

Answer 1

@stamp @jigglypuff314 @Astrophysics @Michele_Laino @pooja195

Answer 2

First we have to find the point of intersection of two curves.

Answer 3

point of intersection: (0.426, 0.653)

Answer 4

@jim_thompson5910

Answer 5

@StudyGurl14 were you able to set up the integral for part (a) ?

Answer 6

no

Answer 7

I'm assuming you have the graph set up since you were able to find the point of intersection?

Answer 8

https://www.desmos.com/calculator/stxakb6pqz

Answer 9

ok great

so notice how the red curve `y = e^(-x)` is above the blue curve `y = sqrt(x)` on the interval from x = 0 to x = 0.426 (approx)

Answer 10

yes

Answer 11

the region between those curves and the y-axis is what we're supposed to find teh area of, right?

Answer 12

let f(x) = e^(-x) and g(x) = sqrt(x)

the area R will be equal to

\[\Large R = \int_{a}^{b}(f(x) - g(x))dx\]

where in this case

a = 0
b = 0.426 (approx)

Answer 13

I agree w/ Jim

Answer 14

\(\Large R = \int_{0}^{0.426}(e^{-x} - \sqrt{x})dx\)

Answer 15

correct, then you can use this rule

\[\Large \int_{a}^{b}(f(x) - g(x))dx=\int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx\]

to break up the integral

Answer 16

and it might help to think of \(\Large \sqrt{x}\) as \(\Large x^{1/2}\) so you can use the power rule