Question

graph the area bounded by f(x) = (-x^2 ) 4x 1 and g(x) = x 1

Answer 1

we have to first find the limits of integration

Answer 2

set both function equal to each other and solve

Answer 3

What is the function? Is it -x^2+4x+1 or -4x+1 or -4x-1? Is it x-1 or x+1?

Answer 4

i have no idea how to use this site

Answer 5

it -x^2+4x+1

Answer 6

is it x+1 or x-1?

Answer 7

malevolence i dont think they mean find the area under curve...graph the area, like shading the bounded area
this is graphing quadratic and linear functions not integration i believe.
nelson is this for calculus?

Answer 8

yea Dumbcow
calc 1 i have a lot more questions if you want to help me

Answer 9

I can help you with anything calculus involved. I just need to know the functions we're dealing with. Is it x+1 or x-1?

Answer 10

the question is graph the area bounded by f(x) = (-x^2 ) 4x +1 and g(x) = x +1

Answer 11

then calculate the area

Answer 12

Okay. You need to set the two equations equal to find the bounds for you integral:
-x^2+4x+1=x+1
-x^2+3x=0
x^2-3x=0
x(x-3)=0
So your bounds are 0,3
Now you need to figure out which curve is on top. Plug in 2 and see. (-4+8+1)=5, 2+1=3
So you have the integral of the top curve minus bottom curve.
\[\int\limits_0^3 (-x^2+4x+1)-(x+1)dx\]
Combine like terms.
Integrate using power rule.
Can you finish it?

Answer 13

ok i was wrong...follow malevolence steps

Answer 14

no i don't know anything

Answer 15

Okay, well combining:
\[\int\limits_0^3(-x^2+3x)dx\]
Then you want to use the power rule that says:
\[\int\limits x^n dx=\frac{x^{n+1}}{n+1}\]
So:
\[\frac{-x^3}{3}+\frac{3x^2}{2}|_0^3=\frac{-1}{3}(3^3-0^3)+\frac{3}{2}(3^2-0)=\frac{-27}{3}+\frac{27}{2}=\frac{9}{2}\]