Help with an integration question needed (1 Viewer)

blackops23

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Hi, I don't quite understand what this question is asking me to do.

Q. Find the area in the first quadrant bounded by y=x^2 and y=x^3 using:
(i) the x-axis
(ii) the y-axis

I understand there is a part enclosed by these two graphs, but does it mean by "using the x/y axes"??

Could some please clarify what I am meant to do in this question?

Help appreciated greatly
 

NewiJapper

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I think it just means finding the area bounded by the two curves in respect to x, meaning you would just use those functions (for the x axis), and then for the y-axis, change the terms around to make it into respect to y. eg x=sqrty and then find the area that way. Of course you would have to find where they intersect first, but that is easy :p
 

funnytomato

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Hi, I don't quite understand what this question is asking me to do.

Q. Find the area in the first quadrant bounded by y=x^2 and y=x^3 using:
(i) the x-axis
(ii) the y-axis

I understand there is a part enclosed by these two graphs, but does it mean by "using the x/y axes"??

Could some please clarify what I am meant to do in this question?

Help appreciated greatly
For the i) part, it's just what you normally do. i.e. find points of intersection, which are x=0,1 and then integrate the difference of the two functions given using these integrands

And for ii) part , turn them into functions of y , which are square and cube roots of y, then solve for points of intersection, you'll get y=0,1 Then integrate the difference of these two functions in terms of x and get the area
 
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SpiralFlex

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Find the intersection points which are obviously (0,0) and (1,1).

i) Area bounded by the two curves using the x axis is given by the formula, Area =

Let's begin to integrate.

Area =

Area = with the limits of 1 and 0.

ii) Area bounded by the two curves using the y axis is given by the formula, Area =

Make x the subject of the formula, and

Area =

Area = with limits of 1 and 0.

Both answers should turn out to be the same, it is just showing you that either way would work.

 
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