A  quartic funtion $ y \equal{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx\plus{}e\ (a\neq 0)$ touches the line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha ,\ \beta \ (\alpha < \beta ).$ Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta$.
1. Given the quartic polynomial function \( y = ax^4 + bx^3 + cx^2 + dx + e \) and the line \( y = px + q \), we know that they touch at \( x = \alpha \) and \( x = \beta \) where \( \alpha < \beta \). This implies that at these points, the function values and their first derivatives are equal.

2. Define the functions:
   \[
   f(x) = ax^4 + bx^3 + cx^2 + dx + e
   \]
   \[
   d(x) = px + q
   \]

3. Since the quartic function touches the line at \( x = \alpha \) and \( x = \beta \), we have:
   \[
   f(\alpha) = d(\alpha) \quad \text{and} \quad f(\beta) = d(\beta)
   \]
   \[
   f'(\alpha) = d'(\alpha) = p \quad \text{and} \quad f'(\beta) = d'(\beta) = p
   \]

4. The difference between the quartic function and the line can be expressed as:
   \[
   |f(x) - d(x)| = a(x - \alpha)^2(x - \beta)^2
   \]

5. The area of the region bounded by these graphs is given by the integral:
   \[
   \sigma [\Gamma_{f,d}] = \int_{\alpha}^{\beta} |f(x) - d(x)| \, dx = a \int_{\alpha}^{\beta} (x - \alpha)^2 (x - \beta)^2 \, dx
   \]

6. To simplify the integral, perform a change of variables. Let \( u = x - \frac{\alpha + \beta}{2} \). Then \( du = dx \) and the limits of integration change from \( \alpha \) to \( -\frac{\beta - \alpha}{2} \) and from \( \beta \) to \( \frac{\beta - \alpha}{2} \).

7. The integral becomes:
   \[
   a \int_{-\frac{\beta - \alpha}{2}}^{\frac{\beta - \alpha}{2}} \left( \left( u + \frac{\beta - \alpha}{2} \right)^2 - \left( \frac{\beta - \alpha}{2} \right)^2 \right)^2 \, du
   \]

8. Simplify the integrand:
   \[
   \left( \left( u + \frac{\beta - \alpha}{2} \right)^2 - \left( \frac{\beta - \alpha}{2} \right)^2 \right)^2 = \left( u^2 - \left( \frac{\beta - \alpha}{2} \right)^2 \right)^2
   \]

9. The integral now becomes:
   \[
   a \left( \frac{\beta - \alpha}{2} \right)^4 \int_{-\frac{\beta - \alpha}{2}}^{\frac{\beta - \alpha}{2}} \left( \left( \frac{2u}{\beta - \alpha} \right)^2 - 1 \right)^2 \, du
   \]

10. Let \( v = \frac{2u}{\beta - \alpha} \), then \( dv = \frac{2}{\beta - \alpha} du \) and the limits of integration change from \( -1 \) to \( 1 \).

11. The integral becomes:
   \[
   a \left( \frac{\beta - \alpha}{2} \right)^5 \int_{-1}^{1} (v^2 - 1)^2 \, \frac{\beta - \alpha}{2} dv
   \]

12. Simplify and evaluate the integral:
   \[
   a \left( \frac{\beta - \alpha}{2} \right)^5 \cdot \frac{\beta - \alpha}{2} \cdot 2 \int_{0}^{1} (v^2 - 1)^2 \, dv
   \]

13. The integral \( \int_{0}^{1} (v^2 - 1)^2 \, dv \) can be computed as:
   \[
   \int_{0}^{1} (v^4 - 2v^2 + 1) \, dv = \left[ \frac{v^5}{5} - \frac{2v^3}{3} + v \right]_{0}^{1} = \frac{1}{5} - \frac{2}{3} + 1 = \frac{8}{15}
   \]

14. Therefore, the area is:
   \[
   2a \left( \frac{\beta - \alpha}{2} \right)^5 \cdot \frac{8}{15} = \frac{a (\beta - \alpha)^5}{30}
   \]

The final answer is \(\boxed{\frac{a (\beta - \alpha)^5}{30}}\)