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secondary 4 | A Maths
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need help with this qn, pls explain too
Date Posted: 2 years ago
Views: 427
To do this, you need to know how the graph looks like first (else, we can’t draw the thing).
The two graphs will intersect at a point defined by
sin x = cos x
sin x / cos x = 1
tan x = 1
We will look at the lowest positive value of x (because the intersection they want is within 0 and pi/2)
x = pi/4
Housing the region x = 0 to x = pi/4 is the curve y = sin x and housing the region x = pi/4 to x = pi/2
Then, area housed by the first region
= INT sin x dx (upper limit pi/4, lower limit 0)
= [-cos x] upper lower
= - cos pi/4 - (-cos 0)
= - (sqrt2) / 2 + 1
Area housed by the second region
= INT cos x dx (upper limit pi/2, lower limit pi/4)
= [sin x] upper lower
= sin pi/2 - sin pi/4
= 1 - (sqrt 2) / 2
Their total area is 2 - sqrt 2.
I let you try the second part first
The two graphs will intersect at a point defined by
sin x = cos x
sin x / cos x = 1
tan x = 1
We will look at the lowest positive value of x (because the intersection they want is within 0 and pi/2)
x = pi/4
Housing the region x = 0 to x = pi/4 is the curve y = sin x and housing the region x = pi/4 to x = pi/2
Then, area housed by the first region
= INT sin x dx (upper limit pi/4, lower limit 0)
= [-cos x] upper lower
= - cos pi/4 - (-cos 0)
= - (sqrt2) / 2 + 1
Area housed by the second region
= INT cos x dx (upper limit pi/2, lower limit pi/4)
= [sin x] upper lower
= sin pi/2 - sin pi/4
= 1 - (sqrt 2) / 2
Their total area is 2 - sqrt 2.
I let you try the second part first
thx :) i managed to complete the qn, however i still cant do qn 6 tho
Where the curve y = -x² + 5x + 4 and the line y = 8 meet,
-x² + 5x + 4 = 8
-x² + 5x - 4 = 0
x² - 5x + 4 = 0
(x - 1) (x - 4) = 0
x = 1 or x = 4
You need this later on, and you will need to realise that y = -x² + 5x + 4 is a sad face graph which will cut the line y = 8 twice, at x = 1 (when the curve is going up) and at x = 4 (when the curve is going down)
-x² + 5x + 4 = 8
-x² + 5x - 4 = 0
x² - 5x + 4 = 0
(x - 1) (x - 4) = 0
x = 1 or x = 4
You need this later on, and you will need to realise that y = -x² + 5x + 4 is a sad face graph which will cut the line y = 8 twice, at x = 1 (when the curve is going up) and at x = 4 (when the curve is going down)
See 2 Answers
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Hope this helps!
Date Posted:
2 years ago
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A possible sketch goes like this.
The area of the region we want is basically the area of the red region in the picture (which is enclosed by the curve and the horizontal line).
The red region is not touching the x-axis directly, so we cannot just take the area under the curve, since the area under the curve from x = 1 to x = 4 is the entire shaded region of both the red and the blue regions (under the curve ONTO the x-axis).
The area which we are looking
= Area under curve - Blue rectangle
= INT (-x² + 5x + 4) dx [upper is x = 4, lower is x = 1] - [3 x 8]
= [-x³/3 + 5x²/2 + 4x] [upper is x = 4, lower is x = 1] - 24
= [-4³/3 + 5 (4)²/2 + 4 (4)] - [-1³/3 + 5 (1)²/2 + 4 (1)] - 24
= [-64/3 + 40 + 16] - [-1/3 + 5/2 + 4] - 24
= 104/3 - 37/6 - 24
= 9/2 units²
The area of the region we want is basically the area of the red region in the picture (which is enclosed by the curve and the horizontal line).
The red region is not touching the x-axis directly, so we cannot just take the area under the curve, since the area under the curve from x = 1 to x = 4 is the entire shaded region of both the red and the blue regions (under the curve ONTO the x-axis).
The area which we are looking
= Area under curve - Blue rectangle
= INT (-x² + 5x + 4) dx [upper is x = 4, lower is x = 1] - [3 x 8]
= [-x³/3 + 5x²/2 + 4x] [upper is x = 4, lower is x = 1] - 24
= [-4³/3 + 5 (4)²/2 + 4 (4)] - [-1³/3 + 5 (1)²/2 + 4 (1)] - 24
= [-64/3 + 40 + 16] - [-1/3 + 5/2 + 4] - 24
= 104/3 - 37/6 - 24
= 9/2 units²
Date Posted:
2 years ago
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