area integration (1 Viewer)

[Arithela]

find the area enclosed by the line y = 2x, the parabola y = -x^2 and the line x =2.

I could do all the ones involving 2 curves, but this question involves 3 curves. How should i do it? Thanks!

 

[Mark576]

Actually, if you draw the graph, it should become readily apparent: First find the area between y=2x, x=2 and the x-axis, easy enough just integrate 2x from x=2 to x=0. Then just find the absolute integral from x=2 to x=0 of -x^2 with respect to x, which will give you the area of the part below the x-axis.

 

[Arithela]

yup, after re-drawing the graph i can see that the area is divided into 2 parts. thanks

also, when sketching, does y = 6x^3 look the same as y = x^3? if so, we show they are different by including a point, yes?

 

[Mark576]

Yeah you could do that, but also make sure to label the graph as y=6x³. Shouldn't be any problem.

 

[Arithela]

thank you. one just more question if you dont mind.

find the indefinite integral with respect to x:

(ax)^1/2 + (ax)^-1/2


I got the answer but I don't know why you need to also bring the 'a' down to the denominator when integrating. Isn't 'a' just another constant?

 

[Arithela]

oh ok.. there wasnt a 'b' there so it confused me. got it now though.

how about this question:

find the area enclosed between the 2 curves:

y = (x + 2)^1/2 and 5y = x + 6

i just spent 30 mins on this question and got the wrong answer

 

[Arithela]

ok... seems like the time system on BOS stuffed up

 

[Mark576]

oh ok.. there wasnt a 'b' there so it confused me. got it now though.

how about this question:

find the area enclosed between the 2 curves:

y = (x + 2)^1/2 and 5y = x + 6

i just spent 30 mins on this question and got the wrong answer

Do you have any trouble drawing the graphs? Draw the graphs and then find the points of intersection, which occur at x=-1 and x=14:

So just find the integral of (x+2)^1/2, evaluated from 14 to -1 and then subtract the area between y = (x+6)/5, the x-axis and the lines x=-1 and x=14 to find the area between the two graphs. Sorry if that was a little muddled, I'm tired, and off to bed.

 

[DJel]

Use the result; the integral of [(ax+b)^n] dx = [(ax+b)^(n+1)]/a(n+1)

So for your question;

(ax)^1/2 + (ax)^-1/2
= [(ax)^3/2]/a(3/2) + [(ax)^1/2]/a(1/2) + C
= 2[root(ax^3)]/3a + 2root(ax)/a + C

 

[nick3157]

i just dropped maths thanks to all this stuff