(Core 3 level) Integrate the function f(x) = 2 -cos(3x) between the bounds 0, pi/3.
f(x) = 2 - cos(3x)integrate function x term2 -> 2x (raise power of x then divide by new power for polynomial functions of x) -cos(3x) -> -(1/3)(sin(3x)) (using a substitution of 3x = u, then cos(u) integrates to sin(u)) Answer:2x - (1/3)(sin(3x)) + csubstituting bounds,(2*(pi/2) - (1/3(sin(3pi/3))) - (20 - (1/3)(sin(30/3)) = 2pi/3 - 0 = 2pi/3.
RG
Related Maths A Level answers
All answers ▸What is the method used for differentiation?
integral of (tan(x))dx using the substitution u = cos(x)
The curve C has a equation y=(2x-3)^5; point P (0.5,-32)lies on that curve. Work out the equation to the tangent to C at point P in the form of y=mx+c
How can you express the complex number z = 2 + 3i in the form z = r(cos x + i sinx)