If `P(x)` is a polynomial of the least degree that has a maximum equal to 6 at `x=1`, and a minimum equalto 2 at `x= 3`,then `int_0^1 P(x)dx` equals:

Options

A. `17/4`
B. `13/4`
C. `19/4`
D. `5/4`

Explanation

To solve the problem step by step, we need to find the polynomial \( P(x) \) that meets the given conditions and then evaluate the definite integral \( \int_0^1 P(x) \, dx \).

### Step 1: Identify the polynomial form
Given that \( P(x) \) has a maximum at \( x = 1 \) and a minimum at \( x = 3 \), we can express \( P(x) \) in terms of its critical points. The polynomial must have roots at these points. Since we need a polynomial of the least degree, we can start with:

\[ P(x) = a(x - 1)^2(x - 3) \]

This form ensures that \( x = 1 \) is a maximum (double root) and \( x = 3 \) is a minimum (single root).

### Step 2: Determine the value of \( a \)
We know that \( P(1) = 6 \) (maximum value). Thus, substituting \( x = 1 \):

\[ P(1) = a(1 - 1)^2(1 - 3) = a(0)(-2) = 0 \]

This does not help us directly, so we need to evaluate \( P(3) \) to find \( a \). We know \( P(3) = 2 \):

\[ P(3) = a(3 - 1)^2(3 - 3) = a(2^2)(0) = 0 \]

Again, this does not help us. Let's find \( P(1) \) and \( P(3) \) in terms of \( a \) and set up equations.

### Step 3: Expand the polynomial
Expanding \( P(x) \):

\[ P(x) = a(x - 1)^2(x - 3) = a((x^2 - 2x + 1)(x - 3)) \]
\[ = a(x^3 - 3x^2 - 2x^2 + 6x + x - 3) \]
\[ = a(x^3 - 5x^2 + 7x - 3) \]

### Step 4: Set up equations for conditions
Now we can use the conditions \( P(1) = 6 \) and \( P(3) = 2 \):

1. For \( P(1) = 6 \):
\[ P(1) = a(1^3 - 5(1^2) + 7(1) - 3) = a(1 - 5 + 7 - 3) = a(0) = 6 \quad \text{(not useful)} \]

2. For \( P(3) = 2 \):
\[ P(3) = a(3^3 - 5(3^2) + 7(3) - 3) = a(27 - 45 + 21 - 3) = a(0) = 2 \quad \text{(not useful)} \]

### Step 5: Find \( a \) using the maximum and minimum
We can directly calculate \( P(1) \) and \( P(3) \) using the polynomial form:

1. At \( x = 1 \):
\[ P(1) = a(1 - 1)^2(1 - 3) + 6 = 6 \]

2. At \( x = 3 \):
\[ P(3) = a(3 - 1)^2(3 - 3) + 2 = 2 \]

### Step 6: Evaluate the integral
Now we can evaluate the integral \( \int_0^1 P(x) \, dx \):

Using the polynomial we derived:
\[ P(x) = a(x^3 - 5x^2 + 7x - 3) \]

Integrate from 0 to 1:
\[ \int_0^1 P(x) \, dx = \int_0^1 (ax^3 - 5ax^2 + 7ax - 3a) \, dx \]
\[ = a\left[\frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 3x\right]_0^1 \]
\[ = a\left(\frac{1}{4} - \frac{5}{3} + \frac{7}{2} - 3\right) \]
\[ = a\left(\frac{1}{4} - \frac{20}{12} + \frac{42}{12} - \frac{36}{12}\right) \]
\[ = a\left(\frac{1}{4} - \frac{14}{12}\right) = a\left(\frac{3}{12} - \frac{14}{12}\right) = a\left(\frac{-11}{12}\right) \]

### Final Calculation
Substituting \( a \) from previous calculations, we can find \( \int_0^1 P(x) \, dx \).