Integrate the equation (p-qt)^n with respect to t
Suppose we had
∫ (2 + 3x)^5 dx
wouldn't that just be (1/5)(1/3)(2+3x)^6 + c ??
so ∫ (p-qt)^n dt = (1/n)(-1/q)(p - qt)^(n+1) + C
1/(n+1) ...
Yup, thanks oobleck for checking me
(1/(n+1) )(-1/q)(p - qt)^(n+1) + C
To integrate the equation (p-qt)^n with respect to t, you can use the substitution method. Let's go through the step-by-step process to solve this integral.
Step 1: Make a substitution
Let u = p - qt, and then differentiate both sides to find du/dt.
du/dt = -q
Step 2: Express dt in terms of du
Rearrange the above equation to solve for dt in terms of du.
dt = -du/q
Step 3: Rewrite the integral in terms of u and du
Substitute u and dt into the original integral equation.
∫ (p-qt)^n dt = ∫ u^n (dt) = ∫ u^n (-du/q)
Step 4: Simplify the integral
Take the constant factor (-1/q) outside the integral.
-1/q ∫ u^n du
Step 5: Integrate with respect to u
Apply the power rule for integration.
-1/q ∫ u^n du = -1/q * (1/(n+1)) * u^(n+1) + C
Step 6: Substitute back the original variable
Replace u with p - qt in the above equation.
-1/q * (1/(n+1)) * (p - qt)^(n+1) + C
Thus, the final integrated equation is:
∫ (p-qt)^n dt = -1/q * (1/(n+1)) * (p - qt)^(n+1) + C