Question

Find the area under the graph of f over the interval [5,8]
8x+3 for x<6
57-x for x>6

Answer 1

it is 8x+3 for x \[8x+3, x \le 6\]

Answer 2

The integral of a function represents the area under it.

\[Area = \int\limits_{5}^{6}(8x+3)dx + \int\limits_{6}^{8}(57-x)dx\]

Answer 3

So then you integrate and solve for a number, and I keep ending with an expression.

Answer 4

i don't understand

Answer 5

A statement :P It should be a number! Idk what im doing wrong

Answer 6

to solve the first part it is F(6) - F(5)

second part [after plus sign]

F(8) - F(6)

Answer 7

\[\int\limits_{5}^{6}(8x+3)dx = 4x^2 + 3x\]\[\int\limits_{6}^{8}(57 - x) dx = 57x - \frac{ x^2 }{ 2 }\]

Answer 8

first part:

\[4(6)^2 + 3(6) - (4(5)^2 + 3(5))\]
seond part:\[57(8) - \frac{ 8^2 }{ 2 } - (57(6) - \frac{6^2}{2})\]

Answer 9

i get 47 + 100 = 147

Answer 10

It's correct! I have no idea what I was doing wrong lol, I think it was just a brain fart.

Answer 11

Thank you so much, highly appreciate it. Nobody likes to answer my questions anymore haha

Answer 12

glad i could help ^_^

might be because the smartscores & medals have been not working for a few days now. i get it demotivated people :P

Answer 13

i guess*

Answer 14

@Euler271 could you quickly do this?\[\int\limits_{0}^{4}(5e^t)dt\]

Answer 15

Im getting 14899 and it's marking wrong

Answer 16

\[5\int\limits_{0}^{4}e^t = 5e^t \left(\begin{matrix}4 \\ 0\end{matrix}\right)= 5 (e^4 - e^0) = 5(e^4 - 1)\]

Answer 17

e^t is the derivative/integral of itself

5 is just a constant so we don't consider it in integration

Answer 18

Right! that's what I got, so why am I getting such a weird number? What is the value of 5(e^4-1)?

Answer 19

267.9907502

but i doubt they want an approximation. you should literally type 5(e^4 - 1) in [if this is online homework]

Answer 20

They got 6699.77 lol :(

Answer 21

http://www.wolframalpha.com/input/?i=integrate+from+0+to+4+5e%5Et

Answer 22

Lol sometimes my online book is wrong..

Answer 23

Area under the graph of f over the interval [-3,3]
2-x^2 if x<0
2 if x>(greater than or equal to) 0

Answer 24

lol yup. textbooks have mistakes sometimes.

i've driven myself crazy because of those.

Answer 25

\[Area = \int\limits_{-3}^{0} (2 - x^2)dx + \int\limits_{0}^{3}2dx\]

Answer 26

I GOT -9 AND IT'S NOT AN ANSWERRRRRRRRRRRRRR

Answer 27

\[2x - \frac{ x^3 }{ 3 }\left(\begin{matrix}0 \\ -3\end{matrix}\right) + 2x \left(\begin{matrix}3 \\ 0\end{matrix}\right)\]\[= 2(0) - \frac{ 0^3 }{ 3 } - (2(-3) - \frac{ (-3)^3 }{ 3 }) + 2(3) - 2(0)\]

area can't be negative ^_^
i'll calculate this 2 sec

Answer 28

LOL not an answer still ;_;
9
3
-15
-6

Answer 29

OH so its 9

Answer 30

:O???

Answer 31

it's 3. my bad

Answer 32

so many negative signs

Answer 33

Okay! I'll trust your work over mine since I'm new to this.
Find the area of the shaded region
f(x)=-x^3-x^2+20x, g(x)=0
[-5,4]

Answer 34

can you make a rough sketch of the shaded region plz?

Answer 35

since some of it is below the x-axis, doing it normally would lead to whatever is below the x-axis being negative (i.e subtracted).

so you must consider them seperately and add them

so you need to do:

\[\int\limits_{-5}^{0}(-x^3-x^2+20x) dx + \int\limits_{0}^{4}(-x^3-x^2+20x) dx\]

the answer:
http://www.wolframalpha.com/input/?i=integrate+-x%5E3-x%5E2%2B20x+from+-5+to+4

you should definitely try it though ^_^

Answer 36

sorry, wolfram is giving you the integral (which can be negative)
however, the AREA itself must be positive, so:

absolute value of this: (positive): http://www.wolframalpha.com/input/?i=integrate+-x%5E3-x%5E2%2B20x+from+-5+to+0

plus this: http://www.wolframalpha.com/input/?i=integrate+-x%5E3-x%5E2%2B20x+from+0+to+4

Answer 37

That's not an option D: lol

Answer 38

-81/12
2137/12
1625/12
-679/12

Answer 39

weird.