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junior college 1 | H2 Maths
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need help w the second part :)
Date Posted: 3 years ago
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In the first part, you would have gotten ½ for the exact area.
In the second part, you are integrating the inverse of the function in the first part, y = sin-¹ x
But we can't integrate this directly since question asks us to use first part's results (it is troublesome to integrate sin-¹x directly anyway)
However, notice the following:
① We can rotate the original function y = sin x such that the y-axis is our independent/horizontal axis and x-axis is our dependent axis/vertical axis.
y = sin x ⇒ x = sin-¹ y
When x = π/3, y = sin π/3 = √3/2
When x = 0, y = sin 0 = 0
② We could then integrate the function x = sin-¹ y with respect to y to find the area bound by y = √3/2, the y-axis and the function.
⇒ ∫₀√³/² sin-¹y dy
This is basically equivalent to ∫₀√³/² sin-¹x dx since we are just swapping the letter y for x
Now, instead of doing any integration, we realise that this area
= (Area of rectangle bounded by the y-axis, x-axis, line y = √3/2 and x = π/3) - ∫₀π/³ sin x dx
So ∫₀√³/² sin-¹x dx
= ∫₀√³/² sin-¹y dy
= Area of rectangle - ∫₀π/³ sin x dx
= (π/3)(√3 / 2 ) - ½
= π/2√3 - ½
= ½(π/√3 - 1)
In the second part, you are integrating the inverse of the function in the first part, y = sin-¹ x
But we can't integrate this directly since question asks us to use first part's results (it is troublesome to integrate sin-¹x directly anyway)
However, notice the following:
① We can rotate the original function y = sin x such that the y-axis is our independent/horizontal axis and x-axis is our dependent axis/vertical axis.
y = sin x ⇒ x = sin-¹ y
When x = π/3, y = sin π/3 = √3/2
When x = 0, y = sin 0 = 0
② We could then integrate the function x = sin-¹ y with respect to y to find the area bound by y = √3/2, the y-axis and the function.
⇒ ∫₀√³/² sin-¹y dy
This is basically equivalent to ∫₀√³/² sin-¹x dx since we are just swapping the letter y for x
Now, instead of doing any integration, we realise that this area
= (Area of rectangle bounded by the y-axis, x-axis, line y = √3/2 and x = π/3) - ∫₀π/³ sin x dx
So ∫₀√³/² sin-¹x dx
= ∫₀√³/² sin-¹y dy
= Area of rectangle - ∫₀π/³ sin x dx
= (π/3)(√3 / 2 ) - ½
= π/2√3 - ½
= ½(π/√3 - 1)
Basic idea is :
y = sin x and y = sin-¹ x are inverses of each other and therefore symmetrical about the line y = x (mirror images of each other)
So, integrating y = sin-¹ x with respect to x from x = 0 to x = √3/2 gives an area that when mapped to the inverse (i.e reflected about the line y = x) , is essentially equivalent to integrating y = sin x (rewritten as x = sin-¹y) with respect to y from y = 0 to y = √3/2
And this area can already be found immediately by subtracting the ∫₀π/³ sin x dx fron the rectangle
y = sin x and y = sin-¹ x are inverses of each other and therefore symmetrical about the line y = x (mirror images of each other)
So, integrating y = sin-¹ x with respect to x from x = 0 to x = √3/2 gives an area that when mapped to the inverse (i.e reflected about the line y = x) , is essentially equivalent to integrating y = sin x (rewritten as x = sin-¹y) with respect to y from y = 0 to y = √3/2
And this area can already be found immediately by subtracting the ∫₀π/³ sin x dx fron the rectangle
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