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C3 applications of integration

The line y=1/3 meets the curve y = 1/x+1 at the point P.
a) Fine the coordinates of P
b) Calculate the area bounded by the line, the curve and y-axis

Done part a) and got (2,1/3)

For part b) I integrated the curve with limits 2 and 0 and got ln3 but answer is ln3 - 2/3
Original post by G.Y
The line y=1/3 meets the curve y = 1/x+1 at the point P.
a) Fine the coordinates of P
b) Calculate the area bounded by the line, the curve and y-axis

Done part a) and got (2,1/3)

For part b) I integrated the curve with limits 2 and 0 and got ln3 but answer is ln3 - 2/3


these are usually problems with log laws.
you should integrate the curve minus the integral of the line since you're looking for the area betweeen the curve and the line
Original post by G.Y
The line y=1/3 meets the curve y = 1/x+1 at the point P.
a) Fine the coordinates of P
b) Calculate the area bounded by the line, the curve and y-axis

Done part a) and got (2,1/3)

For part b) I integrated the curve with limits 2 and 0 and got ln3 but answer is ln3 - 2/3


Sketch a graph, put curve on sketch graph (not 100% accruate btw) then line on there then the straight line y=13 y = \frac{1}{3} which you would then put the points showing on the sketched graph. You would then find the area of the line
(length X width, as it is a square/rectangle) where you would then use to find the area of shaded region it is asking you do find. Be logical when wondering which one to subtract from.
(edited 5 years ago)

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