Step 1: Evaluate the definite integral. The integral to evaluate is $\int_1^p (px-1) dx$. First, find the antiderivative of $px-1$ with respect to $x$: $$ \int (px-1) dx = p\frac{x^2}{2} - x + C $$ Now, apply the limits of integration from $1$ to $p$: $$ \left[ \frac{px^2}{2} - x \right]_1^p = \left( \frac{p(p)^2}{2} - p \right) - \left( \frac{p(1)^2}{2} - 1 \right) $$ $$ = \left( \frac{p^3}{2} - p \right) - \left( \frac{p}{2} - 1 \right) $$ $$ = \frac{p^3}{2} - p - \frac{p}{2} + 1 $$ Combine the terms involving $p$: $$ = \frac{p^3}{2} - \frac{2p}{2} - \frac{p}{2} + 1 $$ $$ = \frac{p^3}{2} - \frac{3p}{2} + 1 $$ Step 2: Set the evaluated integral equal to 0. We are given that $\int_1^p (px-1) dx = 0$. So, we set the expression from Step 1 to 0: $$ \frac{p^3}{2} - \frac{3p}{2} + 1 = 0 $$ Multiply the entire equation by 2 to eliminate the denominators: $$ p^3 - 3p + 2 = 0 $$ Step 3: Solve the cubic equation for $p$. We look for integer roots by testing divisors of the constant term (2), which are $\pm 1, \pm 2$. Test $p=1$: $$ (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0 $$ Since $p=1$ satisfies the equation, $(p-1)$ is a factor of the polynomial. We can use polynomial division or synthetic division to find the other factors. Using synthetic division with root 1: ` 1 | 1 0 -3 2 | 1 1 -2 ---------------- 1 1 -2 0 ` The quotient is $p^2 + p - 2$. So, the cubic equation can be factored as: $$ (p-1)(p^2 + p - 2) = 0 $$ Now, factor the quadratic term $p^2 + p - 2$. We need two numbers that multiply to -2 and add to 1. These numbers are 2 and -1. $$ p^2 + p - 2 = (p+2)(p-1) $$ Substitute this back into the factored cubic equation: $$ (p-1)(p+2)(p-1) = 0 $$ $$ (p-1)^2(p+2) = 0 $$ Step 4: Determine the values of $p$. From $(p-1)^2 = 0$, we get $p-1 = 0$, which means $p=1$. From $(p+2) = 0$, we get $p=-2$. Thus, the values of $p$ are $1$ and $-2$. The final answer is $\boxed{p=1, p=-2}$. Send me the next one 📸