If `int_(0)^(prop)3x^(2)dx=-8`, then the value of `prop` is

Options

A. 0
B. `-2`
C. 2
D. `pm2.`

Answer

To solve the problem, we need to evaluate the definite integral of the function \(3x^2\) from \(0\) to \(\alpha\) and set it equal to \(-8\). Here are the steps to find the value of \(\alpha\):

### Step 1: Set up the integral
We start with the equation given in the problem:
\[ \int_{0}^{\alpha} 3x^2 \, dx = -8 \]

### Step 2: Calculate the integral
To solve the integral, we first find the antiderivative of \(3x^2\):
\[ \int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3 \]
Now we can evaluate the definite integral from \(0\) to \(\alpha\):
\[ \int_{0}^{\alpha} 3x^2 \, dx = \left[ x^3 \right]_{0}^{\alpha} = \alpha^3 - 0^3 = \alpha^3 \]

### Step 3: Set the integral equal to \(-8\)
Now we have:
\[ \alpha^3 = -8 \]

### Step 4: Solve for \(\alpha\)
To find \(\alpha\), we take the cube root of both sides:
\[ \alpha = \sqrt[3]{-8} \]
Since \(-8\) can be expressed as \((-2)^3\), we have:
\[ \alpha = -2 \]

### Conclusion
Thus, the value of \(\alpha\) is:
\[ \boxed{-2} \]

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