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.