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# Integration Find the the shaded area watch

Question 6)

a) -3/x -4x^3/2 + c

bi) Co ordinates of A and B were 1 and 4.

bii) The area under the curve was 4.5 unit^2, but the other smaller area, how do find that?

Also, Question 8) a) A/Centre is (-2,1) and the radius is 5.

but, B) how to do find the intersection point between line L and circle c1.

Please go to the link for the questions.

Thanks to everyone for the help, will rep.
2. Sub x=5 into y=5x -4 -x^2 to find the length of the longer side of the triangle. You already know that the shorter side has length 1 (C-B = 1). Then multiply these lengths together and half it to get the area of the triangle.
3. 6) Since the smaller area is below the x-axis, when you integrate the curve between x=4 and x=5 you must get a negative value.
To find the total area, take the magnitude of this area (i.e. make it positive) and add to the other area.
4. (Original post by Munkle)
Sub x=5 into y=5x -4 -x^2 to find the length of the longer side of the triangle. You already know that the shorter side has length 1 (C-B = 1). Then multiply these lengths together and half it to get the area of the triangle.
No. It's not a triangle - the hypotenuse will be slightly curved.

OP, find the area of the curve between A and B, and then B and C (IIRC correctly the second area will come out as negative because it's under the x axis) ignore any negative signs and add the areas together to get the total.

And for the second one, sub in one equation into the other and solve for x (or y depending on how you substitute) and then use those values by plugging into one of the equations (your choice) to find the corresponding y (or x) value
5. And for 8(b), you know that y= -x + 6 so sub this into the circle equation and simplify. This will give you a value for x, then sub this value back into the same equation to find the corresponding y value.

x^2 + (-x + 6)^2 + 4x - 2(-x + 6) -20 = 0
6. (Original post by Munkle)
Sub x=5 into y=5x -4 -x^2 to find the length of the longer side of the triangle. You already know that the shorter side has length 1 (C-B = 1). Then multiply these lengths together and half it to get the area of the triangle.
It's part of the curve, not the hypotenuse of a right-angled triangle.
7. (Original post by Tempa)

Question 6)

a) -3/x -4x^3/2 + c

bi) Co ordinates of A and B were 1 and 4.

bii) The area under the curve was 4.5 unit^2, but the other smaller area, how do find that?
Integrate the function between 4 and 5 (the magnitude is the area)

Also, Question 8) a) A/Centre is (-2,1) and the radius is 5.

but, B) how to do find the intersection point between line L and circle c1.

Please go to the link for the questions.

Thanks to everyone for the help, will rep.
Solve the equation of circle and the line simultaneously.
(First substitute -x+6 for y in the equation of the circle)
8. (Original post by xXxBaby-BooxXx)
No. It's not a triangle - the hypotenuse will be slightly curved.

OP, find the area of the curve between A and B, and then B and C (IIRC correctly the second area will come out as negative because it's under the x axis) ignore any negative signs and add the areas together to get the total.

And for the second one, sub in one equation into the other and solve for x (or y depending on how you substitute) and then use those values by plugging into one of the equations (your choice) to find the corresponding y (or x) value
So if I sub the values of x for b and c I can get the area above the curve?

That means that integration can get you the areas above the curves as well as below the curves?
9. (Original post by Tempa)
So if I sub the values of x for b and c I can get the area above the curve?

That means that integration can get you the areas above the curves as well as below the curves?
You have to calculate the areas above and below the curve separately and add them together. Integration allows you to find the area between a curve and an axis, doesn't matter if it's above or below

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