Area underneath a curve?

when it intersects the x-axis, it means that y = 0 ----> so (A , 0)

0 = k cos(3 * A)

0 = cos(3A)

π/2 + πn = 3A ---> where n is an integer

(π/6) + (πn/3) = A <----- a)

π/6

∫ ( k cos(3x) - 0 ) dx

0

π/6

∫ k cos(3x) dx

0

. . . . . . . . . π/6

(k/3) sin(3x) ]

. . . . . . .. . 0

(k/3) ( sin(3 * π/6) - sin(3 * 0) )

(k/3) ( sin(π/2) - sin(0) )

(k/3) ( 1 - 0 )

(k/3) ( 1 )

(k/3) or (1/3) * k

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Uhm... properly, in case you have the function A(x) for the section below the curve from a (some consistent) to x (notice that A(a) = 0, because of the fact the section below the curve from a to a is only 0). One only has to tell apart A(x) with know to x, to acquire the unique function. (FTC) as an occasion: the section below some curve from 0 to x could be expressed by using the function g(x) = x ^ 2 + x, so the unique function is f(x) = g'(x) = 2x + a million besides the undeniable fact that, in case you purely understand the fee of the section below the curve from a to b of a few function, you do not have sufficient suggestions to unravel the unique function. e.g: the section below some curve from 0 to a million is two(units), you could have countless opportunities: a million. f(x) = 2 2. f(x) = 4x 3. or maybe piecewise function: f(x) = a million for 0<= x <= 0.5 f(x) = 8x - 3 for x > 0.5 4. ... are you able to get it? :)

y=k cos 3x

Since curve cuts x axis then y = 0

k cos(3x) = 0

k is not zero, therefore cos(3x) =0

or 3x = (2n+1)(pi/2)

x = (2n+1)(pi/6) ...............Ans

Therefor point of A(pi/6,0) ...............ans

when the point is on y axis put x = 0

y = k because cos (0) = 1 and cos (6pi) = -1

then y = -k

then point on y-axis = B(0,k)