IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: halosh92 on March 03, 2010, 01:29:40 pm

Title: c2 edexcel doubt!!!
Post by: halosh92 on March 03, 2010, 01:29:40 pm
sketch the following and find the area of the finite region or regions bounded by the curves and the x-axis :
y=x(x-2)(x-5)
how do u solve such questions...can i have it a step by step method? ???
thx alot.
Title: Re: c2 edexcel doubt!!!
Post by: astarmathsandphysics on March 03, 2010, 02:29:10 pm
The roots are at x=0,2,5

Exapnd the brackets to get y=x^3-7x^2+10x
Integrate beteen the roots in two parts
\int^2_0 x^3-7x^2+10xdx +\int^5_0 -(x^3-7x^2+10x)dx
the minus in the second integral is there because the graph is below the axis
We have
[\frac {x^3}{3}-\frac {7x^3}{3}+ {5x^2}]^2_0 -[\frac {x^3}{3}-\frac {7x^3}{3}+ {5x^2}]^2_0]^5_2 =[\frac {2^3}{3}-\frac {7*2^3}{3}+ {5*2^2}]-0-[\frac {5^3}{3}-\frac {7*5^3}{3}+ {5*5^2}]+[\frac {2^3}{3}-\frac {7*2^3}{3}+ {5*2^2}]=16/3+125/12+16/3=253/12
Title: Re: c2 edexcel doubt!!!
Post by: Saladin on March 03, 2010, 02:36:46 pm
how did you write that fancy math sir?

can u teach me?

like i wanna know how to write it.
Title: Re: c2 edexcel doubt!!!
Post by: halosh92 on March 03, 2010, 04:45:44 pm
The roots are at x=0,2,5

Exapnd the brackets to get y=x^3-7x^2+10x
Integrate beteen the roots in two parts
\int^2_0 x^3-7x^2+10xdx +\int^5_0 -(x^3-7x^2+10x)dx
the minus in the second integral is there because the graph is below the axis
We have
[\frac {x^3}{3}-\frac {7x^3}{3}+ {5x^2}]^2_0 -[\frac {x^3}{3}-\frac {7x^3}{3}+ {5x^2}]^2_0]^5_2 =[\frac {2^3}{3}-\frac {7*2^3}{3}+ {5*2^2}]-0-[\frac {5^3}{3}-\frac {7*5^3}{3}+ {5*5^2}]+[\frac {2^3}{3}-\frac {7*2^3}{3}+ {5*2^2}]=16/3+125/12+16/3=253/12

thx a bunch!!!!!!!!!!! perfect
oh yes true very fancy writings  ;D
Title: Re: c2 edexcel doubt!!!
Post by: astarmathsandphysics on March 03, 2010, 07:10:21 pm
Is latex. Am trying to install it on the wiki. Have been trying 3 days