(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 ▸z = 5 - 3i Find z^2 in a form of a + bi, where a and b are real constants
Consider f(x)=a/(x-1)^2-1. For which a>1 is the triangle formed by (0,0) and the intersections of f(x) with the positive x- and y-axis isosceles?
Prove by contradiction that there is an infinite number of prime numbers.
What is 7 to the power of 8? (