solve P'(t) (11t+11)(t^2+2t+2)^1/3 for the total profit in the first three year

Question

I understand up to the point of : 8(antiderivative sign)with 3 at top and o at the bottom (t+1)(t^2+2t+2)^1/3 dt.

Then my brain freezes and I cant understand what they me to use the value for u and du to intergrate.

Answer 1

Consider the integral of u^1/3: (3u^4/3)/4. This is just the application of the general form of u^n, which when integrated is u^(n+1)/(n+1), where n=1/3. So if we start with u=t^2+2t+2, then du/dt=2t+2=2(t+1); and if Q=(3/4)u^(4/3), then dQ/du=u^(1/3) and dQ/dt=dQ/du*du/dt=u^(1/3)*2(t+1)=2(t+1)(t^2+2t+2)^(1/3). Multiply through by 11/2: (11/2)Q'(t)=(11t+11)(t^2+2t+2)^(1/3). Therefore, (11t+11)(t^2+2t+2)^(1/3)=(11/2)Q'(t)=P'(t).

So Q(t)=(3/4)u^(4/3) (we can ignore the constant of integration because we'll be applying explicit limits). Therefore, P(t)=(11/2)Q(t)=(33/8)(t^2+2t+2)^(4/3).

Now apply limits (t=0 to 3):

33/8[(9+6+2)^(4/3)-2^(4/3)]=33/8(17^(4/3)-2^(4/3))=33/8(43.7118-2.5198)=169.92 approx.