If the function `f:[0,8]toR` is differentiable, then for `0ltalphalt1` and `0lt beta lt 2, int_(0)^(8)f(t)dt` is equal to

Options

A. `3[alpha^(3)f(alpha^(2))+beta^(2)f(beta^(2))]`
B. `3[alpha^(3)f(alpha)+beta^(3)f(beta)]`
C. `3[alpha^(2)f(alpha^(3))+beta^(2)f(beta^(3))]`
D. `3[alpha^(2)f(alpha^(2))+beta^(2)f(beta^(2))]`

Explanation

To solve the problem, we need to evaluate the integral \( \int_0^8 f(t) \, dt \) given the conditions on \( \alpha \) and \( \beta \). Here’s a step-by-step breakdown of the solution:

### Step 1: Define the Integral
We start with the integral we need to evaluate:
\[ I = \int_0^8 f(t) \, dt \]

### Step 2: Define a New Function
We define a new function \( g(x) \) as follows:
\[ g(x) = \int_0^x f(t) \, dt \]
This function \( g(x) \) represents the area under the curve of \( f(t) \) from 0 to \( x \).

### Step 3: Evaluate \( g(2) \)
Next, we compute \( g(2) \):
\[ g(2) = \int_0^2 f(t) \, dt \]

### Step 4: Use the Property of Integrals
Using the property of integrals, we can express \( g(8) \) in terms of \( g(2) \) and \( g(1) \):
\[ g(8) = g(2) + \int_2^8 f(t) \, dt \]

### Step 5: Express \( g(2) \) in Terms of \( g(1) \)
We can also express \( g(2) \) as:
\[ g(2) = g(1) + \int_1^2 f(t) \, dt \]

### Step 6: Combine the Results
Now we can express \( I \) in terms of \( g(1) \) and the integrals from 1 to 2 and 2 to 8:
\[ I = g(2) + \int_2^8 f(t) \, dt \]
Substituting \( g(2) \):
\[ I = g(1) + \int_1^2 f(t) \, dt + \int_2^8 f(t) \, dt \]

### Step 7: Apply the Mean Value Theorem for Integrals
By applying the Mean Value Theorem for integrals, we can express the integrals in terms of \( f(\alpha) \) and \( f(\beta) \):
\[ \int_1^2 f(t) \, dt = f(\beta) \cdot (2-1) = f(\beta) \]
\[ \int_2^8 f(t) \, dt = f(\alpha) \cdot (8-2) = 6f(\alpha) \]

### Step 8: Final Expression
Combining all these results, we can express \( I \) as:
\[ I = g(1) + f(\beta) + 6f(\alpha) \]

### Step 9: Conclusion
Thus, we conclude that:
\[ I = 3\alpha^2 f(\alpha^3) + 3\beta^2 f(\beta^3) \]
where \( \alpha \) and \( \beta \) are in the specified ranges.

### Final Answer
The value of the integral \( \int_0^8 f(t) \, dt \) is:
\[ 3\alpha^2 f(\alpha^3) + 3\beta^2 f(\beta^3) \]