# area under curve help

• Apr 26th 2009, 07:40 AM
Tweety
area under curve help
Quote:

A and B are two points which lie on the curve C, with equation $\displaystyle y = -x^2 + 5x +6$ The diagram shows C and the line l passing through A and B.

(a) Calculate the gradient of C at the point where x=2.
The line l passes through the point with coordinates (2, 3) and is parallel to the tangent to C at the point where x=2.
(b) Find an equation of l.
(c) Find the coordinates of A and B.
The point D is the foot of the perpendicular from B on to the x-axis.
(d) Find the area of the region bounded by C, the x-axis, the y-axis and BD.
(e) Hence find the area of the shaded region.

The equation of L is $\displaystyle y = x+1$

the coordinates of A (-1,0) B (5,6)

for question 'd' I got the area = $\displaystyle \int_{0}^5 [-x^2 + 5x + 6] dx$ = $\displaystyle 50\frac{5}{6}$ which is correct.

I am stuck on question 'e',

If I work out the whole area between point D and A would that be the $\displaystyle \int_{-1}^5 [-x^2 +5x+6 ]$ ?

and than I minus the area of the triangle and the little area above the triangle on the left hand side of the y-axis ?

$\displaystyle \int_{-1}^5 [-x^2 +5x+6 ]$

$\displaystyle (-\frac{x^3}{3} + \frac{5x^2}{2} + 6x )$

$\displaystyle [ -\frac{125}{3} + \frac{125}{2} + 30]$ - $\displaystyle [\frac{1}{3} + \frac{5}{2} - 6]$ = 54 .

$\displaystyle \int_{-1}^0 (-x^2+5x+6)dx$ = $\displaystyle 3\frac{1}{6}$

Area of triangle = $\displaystyle 6 \times 6 \times \frac{1}{2} = 18$

$\displaystyle 18+ 3\frac{1}{6} = 21\frac{1}{6}$

$\displaystyle 54 - 21\frac{1}{6} = 32\frac{5}{6}$

The correct answer is $\displaystyle 33\frac{1}{3}$

Can someone show me where I have gone wrong or whats the correct method?

Thanks.
• Apr 26th 2009, 08:41 AM
curvature
Quote:

Originally Posted by Tweety
The equation of L is $\displaystyle y = x+1$

the coordinates of A (-1,0) B (5,6)

for question 'd' I got the area = $\displaystyle \int_{0}^5 [-x^2 + 5x + 6] dx$ = $\displaystyle 50\frac{5}{6}$ which is correct.

I am stuck on question 'e',

If I work out the whole area between point D and A would that be the $\displaystyle \int_{-1}^5 [-x^2 +5x+6 ]$ ?

and than I minus the area of the triangle and the little area above the triangle on the left hand side of the y-axis ?

$\displaystyle \int_{-1}^5 [-x^2 +5x+6 ]$

$\displaystyle (-\frac{x^3}{3} + \frac{5x^2}{2} + 6x )$

$\displaystyle [ -\frac{125}{3} + \frac{125}{2} + 30]$ - $\displaystyle [\frac{1}{3} + \frac{5}{2} - 6]$ = 54 .

$\displaystyle \int_{-1}^0 (-x^2+5x+6)dx$ = $\displaystyle 3\frac{1}{6}$

Area of triangle = $\displaystyle 6 \times 6 \times \frac{1}{2} = 18$

$\displaystyle 18+ 3\frac{1}{6} = 21\frac{1}{6}$

$\displaystyle 54 - 21\frac{1}{6} = 32\frac{5}{6}$

The correct answer is $\displaystyle 33\frac{1}{3}$

Can someone show me where I have gone wrong or whats the correct method?

Thanks.

Let the "big function" minus the "small one", and integrate from 0 to 5.

Area=$\displaystyle \int_{0}^5 ((-x^2+5x+6)-(x+1))dx$ = $\displaystyle \frac{100}{3} =33\frac{1}{3}$
• Apr 26th 2009, 08:53 AM
Tweety
I dont understand how you got that, becasue although it gives you the right answer is that not just the curve-line expression?

How does that give you the shaded region, as that expression also includes the area thats not shaded?
• Apr 26th 2009, 08:55 AM
Tweety
Actually ignore my other post I get what you did.

Thanks
• Apr 26th 2009, 09:16 AM
curvature
Quote:

Originally Posted by Tweety
I dont understand how you got that, becasue although it gives you the right answer is that not just the curve-line expression?

How does that give you the shaded region, as that expression also includes the area thats not shaded?

Always let the "big function" minus the "small one", and integrate from the left to the right, and you get the area between two curves.