# Maths AS Level Methods Exam Help

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#1
Can someone please help me on this question. I think its on intergration but have no idea where to start. Could someone explain it step by step please for me.

The points P (1,10) and Q (4,4) lie on the curve y= x^3 - 6x^2 + 7x + 8

1) Find the area between the curve and the line PQ
2) Find the equation of the perpendicular bisector of PQ (im thinking maybe this one is to do with midpoints?)

Thanks!

If a diagram is required please let me know and i'll scan it in.
0
16 years ago
#2
(Original post by KaosLisa)
Can someone please help me on this question. I think its on intergration but have no idea where to start. Could someone explain it step by step please for me.

The points P (1,10) and Q (4,4) lie on the curve y= x^3 - 6x^2 + 7x + 8

1) Find the area between the curve and the line PQ
2) Find the equation of the perpendicular bisector of PQ (im thinking maybe this one is to do with midpoints?)

Thanks!

If a diagram is required please let me know and i'll scan it in.
There is no real diagram needed, although one may help you with the questions.

1) Integrate the cure between 1 and 4 and then work out the area enclosed by the line PQ and the x-axis (this will be a trapezium). Then subtract to work out the answer.

2) Yes use mid points and the fact that if a line has gradient a, it's perpendicular bisector has gradient -1/a.

Write back if you need more help.
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#3
How do you know its going to be a trapezium though, thats the part i can never figure out. How do you know what shape its going to be?
0
16 years ago
#4
(Original post by KaosLisa)
How do you know its going to be a trapezium though, thats the part i can never figure out. How do you know what shape its going to be?
If you find the area under a straight line, it's always gonna be a trapezium. Think about it - draw a diagram of a line with some gradient m. Then draw lines from two points on that line to the x-axis. You'll see a trapezium. If the line had gradient 0 (ie was horizontal), you'd have a rectangle.
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16 years ago
#5
(Original post by Nylex)
If you find the area under a straight line, it's always gonna be a trapezium. Think about it - draw a diagram of a line with some gradient m. Then draw lines from two points on that line to the x-axis. You'll see a trapezium. If the line had gradient 0 (ie was horizontal), you'd have a rectangle.
since when was an x^3 graph a straight line?? when you integrate between the points that gives you an approximation of the area under the curve whatever the shape of the curve.
0
16 years ago
#6
since when was an x^3 graph a straight line?? when you integrate between the points that gives you an approximation of the area under the curve whatever the shape of the curve.
the straight line PQ
0
16 years ago
#7
(Original post by KaosLisa)
Can someone please help me on this question. I think its on intergration but have no idea where to start. Could someone explain it step by step please for me.

The points P (1,10) and Q (4,4) lie on the curve y= x^3 - 6x^2 + 7x + 8

1) Find the area between the curve and the line PQ
2) Find the equation of the perpendicular bisector of PQ (im thinking maybe this one is to do with midpoints?)

Thanks!

If a diagram is required please let me know and i'll scan it in.

basicly you need to work from the fformula Y= mx+c

(4-10)/(4-1)=m
then substitute the values in to find c. you have the equation 4 the line. now you integrate under the line and under the curve depending on how the diagram is and then you subtract it. sory if i wasnt of much use to you. take care totsxxx
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