Find the area enclosed by the curve f(x)and the line
2006-09-21 05:44:53 · 4 answers · asked by jess 1 in Science & Mathematics ➔ Mathematics
I have been told the answer is 5/6 but don't know how to reach it.
2006-09-21 05:52:17 · update #1
Since the derivative is f'(x) = 2x+3 then f(x)=x²+3x+c and
y=x²+3x+c is the curve. Find the constant c by substituting either set of points given for x and y and solving for c.
Since the parabola opens 'upwards' (since the x² term is positive) the line is the upper bound of the area, and the parabola is the lower bound of the area.
Now the area between the line and the parabola is given by the integral from -3 to 2 of ((2x+4) - (x²+3x+c)) dx
Have fun ☺
Doug
2006-09-21 05:53:16 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
3
2006-09-21 05:45:48 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Your co-ordinates for P & Q are acceptable yet dy/dx for P is two*4-2=2 no longer 0 so then making use of formulation (y-y1)=m(x-x1) for P y=2x-8 = tangent for Q y=-2x-4 = tangent then the point the position those 2 meet ability the x and y co-ordinates must be equivalent so 2x-8=-2x-4 subsequently 4x=4 x=a million and if x=a million y=2*a million-8 =-6
2016-11-23 13:09:00 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Well...first you will have to integrate f'(x) to get x^2 +3x + C. Now you will need to find C using one of the two points that you have. It will turn out to be -2. So f(x) = x^2+3x -2. Draw a picture to figure out which function is on top and for what limits for x and then integrate the difference of the 2 functions(whatever is on top - whatever is on the bottom) to find the area.
2006-09-21 05:59:25 · answer #4 · answered by nb2020 2 · 0⤊ 0⤋