If `int_0^3 (3ax^2+2bx+c)dx=int_1^3 (3ax^2+2bx+c)dx` where `a,b,c` are constants then `a+b+c=`

Options

A. `a+b+c=3`
B. `a+b+c=1`
C. `a+b+c=0`
D. `a+b+c=2`

Explanation

To solve the problem, we start with the given equation:

\[ \int_0^3 (3ax^2 + 2bx + c) \, dx = \int_1^3 (3ax^2 + 2bx + c) \, dx \]

### Step 1: Rewrite the left-hand side

We can express the left-hand side integral as the sum of two integrals:

\[ \int_0^3 (3ax^2 + 2bx + c) \, dx = \int_0^1 (3ax^2 + 2bx + c) \, dx + \int_1^3 (3ax^2 + 2bx + c) \, dx \]

### Step 2: Set the equation

Now we can set the two integrals equal to each other:

\[ \int_0^1 (3ax^2 + 2bx + c) \, dx + \int_1^3 (3ax^2 + 2bx + c) \, dx = \int_1^3 (3ax^2 + 2bx + c) \, dx \]

### Step 3: Cancel the common integral

Since the integral from \(1\) to \(3\) appears on both sides, we can cancel it out:

\[ \int_0^1 (3ax^2 + 2bx + c) \, dx = 0 \]

### Step 4: Compute the integral from 0 to 1

Now we compute the integral:

\[ \int_0^1 (3ax^2 + 2bx + c) \, dx = \left[ ax^3 + bx^2 + cx \right]_0^1 \]

### Step 5: Evaluate at the limits

Evaluating at the limits \(0\) and \(1\):

\[ = \left( a(1)^3 + b(1)^2 + c(1) \right) - \left( a(0)^3 + b(0)^2 + c(0) \right) \]

This simplifies to:

\[ = a + b + c - 0 = a + b + c \]

### Step 6: Set the equation to zero

Since we have established that the integral equals zero:

\[ a + b + c = 0 \]

### Conclusion

Thus, the final result is:

\[ a + b + c = 0 \]