f(x) = int(sqrt(x+3) - x)dxf(x)=∫(√x+3−x)dx
Split the integral into two separate integrals:
f(x) = int(sqrt(x+3) )dx - int(x)dxf(x)=∫(√x+3)dx−∫(x)dx
u-substitution is your best friend
let u = x+3u=x+3
let du = 1 dxdu=1dx
let dx = dudx=du
find the integral by substituting u into the square root and simplify:
f(x) = int(u^(1/2))du -int(x)dxf(x)=∫(u12)du−∫(x)dx
f(x) = (u^(1/2 + 1))/(3/2) - x^2 / 2 + C f(x)=u12+132−x22+C
f(x) = u^(3/2) / (3/2) - 1/2 x^2 + C f(x)=u3232−12x2+C
f(x) = 2/3u^(3/2) - 1/2 x^2 + C f(x)=23u32−12x2+C
Plug x+3 back in for u:
f(x) = 2/3(x+3)^(3/2) - 1/2 x^2 + C f(x)=23(x+3)32−12x2+C
plug in f(1)=-4f(1)=−4 to find the c-value
-4 = 2/3(1+3)^(3/2) - 1/2 (1)^2 + C −4=23(1+3)32−12(1)2+C
-4 = 2/3(4)^(3/2) - 1/2 + C −4=23(4)32−12+C
-7/2 = 16/3 +C−72=163+C
C=-53/6C=−536
plug in the c-value into your integrated equation
f(x) = 2/3(x+3)^(3/2) - 1/2 x^2 + Cf(x)=23(x+3)32−12x2+C
f(x) = 2/3(x+3)^(3/2) - 1/2 x^2 - 53/6 f(x)=23(x+3)32−12x2−536
