# Evaluate the area of the surface between the graph of the function f(x) = cosx +x + x^2/12 and the lines x=0, x=pi/2, x axis

*print*Print*list*Cite

To find the area between f(x) and x1 and x2, we need first to determine the integral for f(x)

f(x) =cosx + x + x^2/12

intg f(x) = intg [cosx + x + x^2/12]\

= intg(cosx) + intg(x) + intg (x^2/12)

= sinx + x^2/2 + (1/12)*x^3/3

Now we need to determine the area between x1= 0 and x2= pi/2

==> intg(f(x2) - intg(f(x1) = (sinpi/2) + (pi/2)^2 + (1/3*12)*(pi/2)^3 - [ sin0 + 0 + 0]\

= 1 + (pi/2)^2 + (1/36)*(pi/2)^3 - 0

==> the area isL 1+ pi^2/4 + pi^3/ 288

The area A of a function f(x) between x = a and x =b is given by:

A = Integral f(x) dx fro x= a to x = b.

f(x) = cosx+x+x^2/12. bounded by x=0 and x pi/2

A = Integral {cosx+x+x^2/12} dx = 0 to x =pi/2.

= { [sinx+x^2/2 +x^3/3*12] at x= pi/2 } - { [sinx+x^2/2 +x^3/3*12] at x= 0 }

= sin pi/2 +[(p/2)^2 ]/2 + (p/2)^3 / 36

= 1 +pi^2/8+pi^3/288

To evaluate the surface area between f(x), x-axis, x = 0 and x = pi/2, we'll have to calculate the integral, where the integrand is represented by f(x) and the calculus limits for the integrand are x=0 and x=pi/2.

We'll use also the property of integral addition and we'll calculate the integral of f(x) as a sum of integrals of each term of f(x).

We'll consider the terms of f(x) as being functions: f1(x),f2(x), f3(x).

f(x)=f1(x)+f2(x)+f3(x)

Integral f(x)=Integralf1(x)dx+Integralf2(x)dx+Integralf3(x)dx

Integral f1(x)=Integral cos x= sin x= sin pi/2 - sin 0 = 1-0 = 1

Integral f1(x)= 1

Integral f2(x)=Integral x= x^2/2=pi^2/8-0^2/2=(pi^2)/8

Integral f3(x)=Integral x^2/12=1/12*Integral x^2

Integral f3(x)==(1/12)*(x^3)/3=1/12*(pi^3/24)

**Integral f(x)=1+(pi^2)/8+(pi^3/288)**