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secondary 4 | A Maths

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Need help with question 13. It requires you to somehow find the value without applying any integration rules. Topic is area of a region. Source: Additional Maths 360. All help appreciated!

1 represents the length of the hypotenuse of a right angled triangle.

x represents the length of 1 side of it, and it's part of the x axis.

y represents the length of the third side of the triangle, and is parallel to the y axis.

Using Pythagoras theorem,

x² + y² = 1²

y² = 1 - x²

y = ± √(1 - x²)

We only consider the positive part as the negative part is another semicircle , a reflection of this semicircle along the x axis.

So the hypotenuse length is fixed at 1 but the x and y coordinates are variable.

You'll see that all the coordinates on the graph form a semicircle with radius of 1 unit.

From x = 0 to x = 1, it's a quarter circle.

Area of quarter circle = ¼π(1units)²

= ¼π units²

The line y = 1 - x forms a right angled isosceles triangle with the x and y axis, with x intercept 1 and y intercept 1.

Area of this triangle is = ½ x 1 x 1 = ½ units²

Required area

= Area of quarter circle - area of triangle

= ¼π units² - ½ units²

= ¼ (π - 2) units²

You are essentially finding a small half-leaf/segment

The two semicircles form a circle with centre (0,0) and radius 1

From x² + y² = 1²

(x - 0)² + (y - 0)² = 1²

The straight line will form a right angled triangle with the x-axis.

From there we just calculate the integral by the sum/difference of areas.

y = √(1 - x²) would need some inference from the Pythagoras' theorem or circle equation. They are linked.

Technically not the same as circle, but derived from it.

Anyway it's a H.O.T corner question so it's more of a higher order thinking question that goes beyond the mere usual applying of integration/expand the student's horizons/connecting the dots.

It won't come out in national exams

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I have included the graph of the curve (in red), the straight line (in blue) and an additional graph below the x-axis (in green) to illustrate J's point that the equation of the curve is actually that of a (semi) circle with the origin (0, 0) as its centre and radius 1 unit.

To describe in words, the integral gives the area of the region bounded by the curve y = sqrt (1 - x^2) and the line y = 1 - x.

As we well know it, this area is simply the area of the quarter circle (quadrant) with radius 1 unit minus the area of the triangle.

Value of integral

= Area of quadrant - Area of triangle

= 1/4 x Pi x 1 x 1 - 1/2 x 1 x 1

= (Pi/4 - 1/2) units^2