Find the area R under the curve when f(x)=xcos(x) between the limits x=0 and x=2

After careful consideration, we can see the function f(x) is the product of two individual functions. Usually, to find the area under a curve we integrate its function between a set limit. Given our situation, to integrate a function of this nature we use a rule known as integration by parts. The formula for this is as follows:  intergral[u*(dv/dx)] = uv-intergral[(du/dx)*v]      (eqn1). Our first step to tackle this problem is to split the function into two: u=x         dv/dx=cos(x) We differentiate u=x to get du/dx=1 and integrate dv/dx=cosx to give us v=sin(x). With the expressions obtained substitute it into the equation above (eqn1). we get: R=xsin(x)-intergral[sin(x)] R=xsin(x)+cos(x) To find the total area we subtitute within the given limits x=0 and x=2. To get: R=[xsin(x)+cos(x)]₀²  R=2sin(2)+cos(2)−1 R=0.402 REMEMBER: To always use radians when dealing with trignometric functions in calculus.