Question

find the area of the region between y=x, y=x^2+1, x=-1 x=2

Answer 1

hi yeah i said the question wrong last time sorry lol.

Answer 2

ok... so same thing


\[\int\limits_{-1}^{0} (x^2 + 1)dx + \frac{1}{2} \times 1 \times 1 + \int\limits_{0}^{2}(x^2 + 1)dx - \int\limits_{0}^{2} (x) dx\]

that would be my solution

Answer 3

hmm. okay cause we were never told to find areas of shapes in this lesson.

Answer 4

thank you tho appreciate your help!

Answer 5

ok... so you for the area below the x-axis its

\[\left| \int\limits_{-1}^{0} (x) dx\right|\]

if you calculate the area below the axis is will be a negative... so to eliminate that problem have the integral inside absolute value symbols

Answer 6

I would have given you extra marks for using a simple method for finding an area...

Answer 7

well we were taught the vertical rectangle method so i think if i use that the answer will be different than your method ? D;

Answer 8

im going to see what i come up with hang on.

Answer 9

here is a geogebra worksheet with the information move the slider to get it to work.

you may have to download the free software... it's worth it

I have also attached the image

Answer 10

I got the answer to be 9/2..

Answer 11

the method you showed me gave me 11/6 -.-

Answer 12

what should i do lol.

Answer 13

can i show you the method i did ?

Answer 14

2
∫ x^2 + 1 - x dx =
-1
...........................2
x^3/3 + x - x^2/2 | =
..........................-1

8/3 + 2 - 2 - (-1/3 - 1 - 1/2) =
8/3 + 1/3 + 1 + 1/2 =
10 1/2

Answer 15

oops thats supposed to be an integral

Answer 16

ok... the value is 4.5

so this is what I did

\[\int\limits_{-1}^{0} (x^2 + 1)dx = [\frac{1}{3}x^3 + x]^{0}_{-1} = \frac{4}{3}\]

and

\[\left| \int\limits_{-1}^{0} (x) dx\right| = [\frac{x^2}{2}]^0_{-1} = \frac{1}{2}\]

next its

\[\int\limits_{0}^{2}(x^2 + 1)dx = [\frac{1}{3} x^3 + x]^{2}_{0} = \frac{8}{3}\]

and lastly

\[\int\limits_{0}^{2}(x) dx = [\frac{1}{2} x^2]^2_{0} = 2\]

so the area between the curves is

\[A = \frac{4}{3} + \frac{1}{2} + (\frac{8}{3} + 2) - 2\]

I think you'ff find its

\[\frac{27}{6}...or... 4\frac{1}{2}\]

Answer 17

hope it helps

Answer 18

Thank you it does!!

Answer 19

Actually you know what we both got the same answer with different method, i just integrated wrong, Lmao it is 4.5 thank you!