Given a line segment $PQ$ with endpoints on the parabola $y=x^2$ such that the area bounded by $PQ$ and the parabola always equal to $\frac 43.$ Find the equation of the locus of the midpoint $M$.
1. Let $\alpha$ and $\beta$ be the $x$-coordinates of points $P$ and $Q$ on the parabola $y = x^2$, with $\alpha < \beta$. The line $PQ$ intersects the parabola at these points, so we can write the equation of the line $PQ$ as $y = l(x)$.

2. The equation of the line $PQ$ can be expressed as $l(x) - x^2 = -(x - \alpha)(x - \beta)$. This is because the line $PQ$ intersects the parabola at $x = \alpha$ and $x = \beta$.

3. The area bounded by the line $PQ$ and the parabola $y = x^2$ is given by:
   \[
   \text{Area} = \int_{\alpha}^{\beta} (l(x) - x^2) \, dx
   \]
   Given that this area is always equal to $\frac{4}{3}$, we have:
   \[
   \frac{4}{3} = -\int_{\alpha}^{\beta} (x - \alpha)(x - \beta) \, dx
   \]

4. To evaluate the integral, we expand the integrand:
   \[
   (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta
   \]
   Thus,
   \[
   \int_{\alpha}^{\beta} (x - \alpha)(x - \beta) \, dx = \int_{\alpha}^{\beta} (x^2 - (\alpha + \beta)x + \alpha\beta) \, dx
   \]

5. Evaluating the integral term by term:
   \[
   \int_{\alpha}^{\beta} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{\alpha}^{\beta} = \frac{\beta^3}{3} - \frac{\alpha^3}{3}
   \]
   \[
   \int_{\alpha}^{\beta} x \, dx = \left[ \frac{x^2}{2} \right]_{\alpha}^{\beta} = \frac{\beta^2}{2} - \frac{\alpha^2}{2}
   \]
   \[
   \int_{\alpha}^{\beta} 1 \, dx = \left[ x \right]_{\alpha}^{\beta} = \beta - \alpha
   \]

6. Combining these results:
   \[
   \int_{\alpha}^{\beta} (x^2 - (\alpha + \beta)x + \alpha\beta) \, dx = \left( \frac{\beta^3}{3} - \frac{\alpha^3}{3} \right) - (\alpha + \beta) \left( \frac{\beta^2}{2} - \frac{\alpha^2}{2} \right) + \alpha\beta (\beta - \alpha)
   \]

7. Simplifying the expression:
   \[
   \frac{\beta^3 - \alpha^3}{3} - \frac{\alpha + \beta}{2} (\beta^2 - \alpha^2) + \alpha\beta (\beta - \alpha)
   \]
   \[
   = \frac{\beta^3 - \alpha^3}{3} - \frac{\alpha + \beta}{2} (\beta - \alpha)(\beta + \alpha) + \alpha\beta (\beta - \alpha)
   \]
   \[
   = \frac{\beta^3 - \alpha^3}{3} - \frac{(\alpha + \beta)^2 (\beta - \alpha)}{2} + \alpha\beta (\beta - \alpha)
   \]
   \[
   = \frac{\beta^3 - \alpha^3}{3} - \frac{(\alpha + \beta)^2 (\beta - \alpha)}{2} + \alpha\beta (\beta - \alpha)
   \]

8. Given that the area is $\frac{4}{3}$, we have:
   \[
   \frac{4}{3} = \frac{1}{6} (\beta - \alpha)^3
   \]
   \[
   \beta - \alpha = 2
   \]

9. Let $M(X, Y)$ be the midpoint of $PQ$. Then:
   \[
   X = \frac{\alpha + \beta}{2}
   \]
   \[
   Y = \frac{\alpha^2 + \beta^2}{2}
   \]

10. Using the identity $(\alpha + \beta)^2 + (\alpha - \beta)^2 = 2(\alpha^2 + \beta^2)$, we have:
    \[
    (2X)^2 + 2^2 = 2(2Y)
    \]
    \[
    4X^2 + 4 = 4Y
    \]
    \[
    Y = X^2 + 1
    \]

11. Therefore, the equation of the locus of the midpoint $M$ is:
    \[
    \boxed{y = x^2 + 1}
    \]