To find the numberical value of `int_(-2)^(2) (px^(3)+qx+s)dx` it is necessary to know the values of the constants:

Options

A. p
B. q
C. s
D. p and s

Answer

To solve the integral \( I = \int_{-2}^{2} (px^3 + qx + s) \, dx \) and determine which constants are necessary to find its numerical value, we can follow these steps:

### Step 1: Break down the integral
We can separate the integral into three parts:
\[ I = \int_{-2}^{2} px^3 \, dx + \int_{-2}^{2} qx \, dx + \int_{-2}^{2} s \, dx \]

### Step 2: Analyze the first integral \( \int_{-2}^{2} px^3 \, dx \)
The function \( f(x) = px^3 \) is an odd function because:
\[ f(-x) = p(-x)^3 = -px^3 = -f(x) \]
For odd functions, the integral over symmetric limits around zero is zero:
\[ \int_{-2}^{2} px^3 \, dx = 0 \]

### Step 3: Analyze the second integral \( \int_{-2}^{2} qx \, dx \)
Similarly, the function \( g(x) = qx \) is also an odd function:
\[ g(-x) = q(-x) = -qx = -g(x) \]
Thus, this integral is also zero:
\[ \int_{-2}^{2} qx \, dx = 0 \]

### Step 4: Analyze the third integral \( \int_{-2}^{2} s \, dx \)
The function \( h(x) = s \) is a constant function, which is an even function:
\[ h(-x) = s = h(x) \]
For even functions, we can use the property of integration:
\[ \int_{-a}^{a} h(x) \, dx = 2 \int_{0}^{a} h(x) \, dx \]
So, we have:
\[ \int_{-2}^{2} s \, dx = 2 \int_{0}^{2} s \, dx = 2s \cdot (2 - 0) = 4s \]

### Step 5: Combine the results
Now, combining all parts, we find:
\[ I = 0 + 0 + 4s = 4s \]

### Conclusion
To find the numerical value of the integral \( I \), we need to know the value of the constant \( s \). Therefore, the answer is that it is necessary to know the value of \( s \).