integration area question (1 Viewer)

[velox]

just factorise it to find the x-intercepts so y=(1-2x)(x+3) i.e y=0 when x=-3, 1/2 so you just integrate y = -2x²-5x+3 between -3 and 1/2

{-3-->1/2} ∫-2x²-5x+3 dx
=[-2/3.x³ -5/2.x² +3x]{-3-->1/2)
= [19/24 - (-13 1/2)]
=14 7/24

I hope I worked that right but that's the gist of it.

Your probably going to be pissed off now Cos i wrote out the wrong question. Sorry!
It was supposed to be:

Find the area enclosed between the curve y = x<sup>2</sup?>-2x- 3 amd the x axis.
I keep gettign 12 2/3 but its 10 2/3. Not sure what i did wrong.

 

[KFunk]

Your probably going to be pissed off now Cos i wrote out the wrong question. Sorry!
It was supposed to be:

Find the area enclosed between the curve y = x²-2x- 3 amd the x axis.
I keep gettign 12 2/3 but its 10 2/3. Not sure what i did wrong.

Haha, dude, learn to write the question right .

y = x²-2x- 3
= (x-3)(x+1) i.e integrate between -1 and 3

{-1-->3} &int; x²-2x- 3
=[1/3.x³ -x² -3x] {-1-->3}
=[-9 - (1 2/3)] = -10 2/3
hence the area under the curve is 10 2/3 units²

 

[velox]

Thanks. Silly mistakes are plaguing my work lol

 

[m_isk]

factorise x^2-2x-3 and you get (x-3)(x+1)... Again a graph will be useful. The are required is that which is above the curve but below the x axis. therefore, you will need to integrate between 3 and -1.
Integral x^2-2x-3
=[(x^3/3)-x^2-3x] where [] means between 3 and -1.
={(27/3 -9-9)-(-1/3-1+3)
=-10 2/3.
Note: the answer is negative because the area we are required to find is BELOW the x axis...if we were required to simply INTEGRATE, then -10 2/3 is an acceptable answer. However, since we are required to find the area, and are cant be negative duh, then the area is the absolute value of -10 2/3 which is 10 2/3 and dont forget the units squared!!

 

[m_isk]

hey KFunk where'd that come from?!?!?! ah well now velo's got two solutions to stare at!!

 

[velox]

factorise x^2-2x-3 and you get (x-3)(x+1)... Again a graph will be useful. The are required is that which is above the curve but below the x axis. therefore, you will need to integrate between 3 and -1.
Integral x^2-2x-3
=[(x^3/3)-x^2-3x] where [] means between 3 and -1.
={(27/3 -9-9)-(-1/3-1+3)
=-10 2/3.
Note: the answer is negative because the area we are required to find is BELOW the x axis...if we were required to simply INTEGRATE, then -10 2/3 is an acceptable answer. However, since we are required to find the area, and are cant be negative duh, then the area is the absolute value of -10 2/3 which is 10 2/3 and dont forget the units squared!!

yeh i missed the minus sign on the 1 in the 2^ND substitution which resulted in the answer being 12 2/3 instead of 10 2/3. Thanks.