Integration help needed C2

Original post by bigmansouf

My question;

For the curve and straight lines, find their points of intersection, draw a sketch on the same axes and find the area of segment cut off from each curve by the corresponding straight line.

a) y=(x+1(x-2), x-y+1=0

sOLUTION
points of coordinates (-1,0) (3,4)

Area required = Area of trianagle -

\int_{3}^{2} x^{2}-x-2

=

(\frac{1}{2}\times 4 \times4 )- \int_{3}^{2} x^{2}-x-2

=

8 - \int_{3}^{2} x^{2}-x-2

break into two

\int_{2}^{3} x^{2}-x-2 = [\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x]_{2}^{3}

=(\frac{3^{3}}{3}-\frac{3^{2}}{2}-2(3))-(\frac{2^{3}}{3}-\frac{2^{2}}{2}-2(2))

= -\frac{3}{2} - (-\frac{10}{3}) = \frac{11}{6}

bring it together
area required =

8-\frac{11}{6} = \frac{37}{6}

please help me and point out what i did wrong
the book says the answer is

\frac{32}{3}

please help me understand where i went wrong
i have attached my sketch

I also got 32/3 by integrating x+1 -(x+1)(x-2) between -1 and 3

Reply 2

Original post by bigmansouf

My question;

For the curve and straight lines, find their points of intersection, draw a sketch on the same axes and find the area of segment cut off from each curve by the corresponding straight line.

a) y=(x+1(x-2), x-y+1=0

sOLUTION
points of coordinates (-1,0) (3,4)

Area required = Area of trianagle -

\int_{3}^{2} x^{2}-x-2

=

(\frac{1}{2}\times 4 \times4 )- \int_{3}^{2} x^{2}-x-2

=

8 - \int_{3}^{2} x^{2}-x-2

break into two

\int_{2}^{3} x^{2}-x-2 = [\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x]_{2}^{3}

=(\frac{3^{3}}{3}-\frac{3^{2}}{2}-2(3))-(\frac{2^{3}}{3}-\frac{2^{2}}{2}-2(2))

= -\frac{3}{2} - (-\frac{10}{3}) = \frac{11}{6}

bring it together
area required =

8-\frac{11}{6} = \frac{37}{6}

please help me and point out what i did wrong
the book says the answer is

\frac{32}{3}

please help me understand where i went wrong
i have attached my sketch

what area are you finding?

Reply 3

OP

Original post by TeeEm

what area are you finding?

i assume that the area i am asked to find is the one coloured in the pic

thanks for the help

Reply 4

Original post by bigmansouf

i assume that the area i am asked to find is the one coloured in the pic

thanks for the help

I think it is that plus the bit under the x axis, bounded by the curve.

Reply 5

OP

Original post by TeeEm

I think it is that plus the bit under the x axis, bounded by the curve.

in that sense would you say that my method was wrong

Reply 6

OP

Original post by TeeEm

I also got 32/3 by integrating x+1 -(x+1)(x-2) between -1 and 3

thanks found this answer it seems that i was wrong
i was not looking for the right area

i think if it was suppose to be the area that i thought it would have said the curve, straight line and x-axis

thank i understand where i was going wrong

(edited 8 years ago)

Reply 7

Original post by bigmansouf

in that sense would you say that my method was wrong

no
you just need to add the bit under

integrate the curve between -1 and 2
turn the answer positive and add it to 37/6 and see what you get
(I did it differently)

Reply 8

OP

Original post by TeeEm

no
you just need to add the bit under

integrate the curve between -1 and 2
turn the answer positive and add it to 37/6 and see what you get
(I did it differently)

your way would be the best and simplest method but i decide to apply the alternative of using areas of shapes and then subtracting it i thought it would be easier
in most cases it is but not in this one

Reply 9