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Integration

Find the shaded area enclosed by the curve y=2 squareroot x, the line x+y=8 and the x axis.
So the points would be x=0 and x=8. When I try to find the area using integration iam not able to get the ans. Pls help
Reply 1
Original post by Shas72
Find the shaded area enclosed by the curve y=2 squareroot x, the line x+y=8 and the x axis.
So the points would be x=0 and x=8. When I try to find the area using integration iam not able to get the ans. Pls help

15723567371011667843292606690414.jpg
Original post by Shas72
Find the shaded area enclosed by the curve y=2 squareroot x, the line x+y=8 and the x axis.
So the points would be x=0 and x=8. When I try to find the area using integration iam not able to get the ans. Pls help


Post your working
Original post by Shas72
15723567371011667843292606690414.jpg

Find the x-value where the curve and the line cross.
Work out the area in two parts.
An integration for the area under the curve on the left.
The area on the right is a triangle so that doesn't need integration.
Reply 4
Original post by RDKGames
Post your working

15723571101721314211548092458763.jpg
Original post by Shas72
15723571101721314211548092458763.jpg

Your working is for finding all of the area under the curve, not just the shaded part.
Reply 6
I did not get your point. the first coordinate is (0,0) and the other one would be (8,0) right?
Answer.jpg

In my diagram a is the x value where the curve and the line intersect. You need to solve an equation to find the value of a.
Then integrate (like you did) from 0 to a. That finds the area to the left of my red line.
Then add the area of the triangle that is to the right of my red line.

Original post by Shas72
I did not get your point. the first coordinate is (0,0) and the other one would be (8,0) right?
Reply 8
I got it. thanks. the answer will be 18 and 2/3 units^2
i waz u i would answer 5 always works for meeeeeeeeeeee

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