Evaluate X 4 Dx Using A Geometric Formula 6 0 Verifying Results in Geometry Using Calculus

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Verifying Results in Geometry Using Calculus

The formulas used in geometry are as follows:

Area of ​​triangle = 0.5 * base * height

Circumference of circle = 2 * pi * r

Area of ​​a circle = pi * r * r

Let’s prove each of these using calculus, first proving the area of ​​the triangle.

Taking a scalene triangle ABC, we draw a straight line from one of the vertices to the opposite side, so that it meets the opposite side at right angles to the base. In summary we have divided the original triangle into two right triangles.

Let us derive an expression for the area of ​​a triangle using the equation of a straight line as Y = m * X + C; First, let’s define the area of ​​a triangle as the sum of the two areas enclosed by the two sides of the triangle, the altitude and the base. Elevation cuts the base into two pieces.

We take a scalene triangle whose height is 5 and base is 10. Let the point of intersection of the altitude with the base be the origin or (0,0). If the three vertices are A, B and C respectively, then if the intersection of point A and line BC is at point D, then the coordinates of point A are (0.5), point B is (4,0) and point C is (-6, 0) Yes. The length of the three sides is BC = 10; AC = sqrt(61) equals about 7.8 and AB = sqrt(41) equals about 6.4.

These three points form a triangle as AB + BC > AC; AC + BC > AB; AC + AB > BC. Also, since three sides of a triangle are not equal, the triangle is scalene.

The area of ​​this triangle is 0.5 * base * height = 0.5 * 10 * 5 = 25 square units.

Let us prove this using integral calculus.

The slope of side AB is -tan(ABD) or -tan(ABC) which is equal to -1.25. The equation of line AB is Y = 5 – 1.25X. Similarly the equation of line AC is Y= 5X/6 + 5. Let both these expressions be integrated between suitable limits.

The area bounded by line AB is the integral (Ydx) between x = 0 and x = 4.

Integral(Ydx) is nothing but 5X – 1.25 * X * X /2. The area between the limits is 20 – 10 = 10.

Similarly the area bounded by the line AC and the origin is Integral(5x/6 + 5)dx which is equal to 5*X*X/12 + 5*X. The limits of integration are 0 to -6 which evaluates to 15 -30 or -15. Taking the absolute value, the sum of both the fields is 15 + 10 which equals 25. This is easily avoided with an analytical expression for the area of ​​a triangle that evaluates to 25 square units.

Let’s derive an expression for the circumference of a circle. If one considers a small area of ​​field d (theta). The area of ​​the region will be r * sin(d(theta)). This can be approximated to r*d(theta) for small values ​​of d(theta). If one integrates this expression between 0 and 2 * pi, it works out to 2 * pi * r. This expression is the quantity of the known expression for the circumference of a circle.

Similarly let us proceed to evaluate the expression for the area of ​​a circle. Let’s calculate the length of an arc subtended by a small angle (theta). The perimeter is nothing but r * d (theta). The area of ​​the sector can be approximated to the area of ​​the triangle as 0.5 * r * r * * d (theta). Integrating 0.5 * r * r * d(theta) between 0 and 2 * pi, evaluates to pi * r * r which fits the expression for the area of ​​the triangle.

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