The value of ` int_(0)^(x)[t+1]^(3) dt` (where, [.] denotes the greatest integer function of x) is qeual to

Options

A. `(([x]([x]+1))/(2))^(2)+([x]+1)^(3){x}`
B. `(([x]([x]+1))/(2))^(3)+([x]+1)^(3){x}`
C. `(([x]([x]+1))/(2))^(3)+([x]+1)^(2){x}`
D. None of these

Answer

To solve the integral \( \int_{0}^{x} (t + 1)^3 \, dt \), where \( [x] \) denotes the greatest integer function of \( x \), we will break down the solution step by step.

### Step 1: Expand the integrand
First, we need to expand the integrand \( (t + 1)^3 \):
\[ (t + 1)^3 = t^3 + 3t^2 + 3t + 1 \]

### Step 2: Set up the integral
Now, we can set up the integral:
\[ \int_{0}^{x} (t + 1)^3 \, dt = \int_{0}^{x} (t^3 + 3t^2 + 3t + 1) \, dt \]

### Step 3: Integrate term by term
Next, we will integrate each term separately:
\[ \int_{0}^{x} t^3 \, dt = \left[ \frac{t^4}{4} \right]_{0}^{x} = \frac{x^4}{4} \]
\[ \int_{0}^{x} 3t^2 \, dt = 3 \left[ \frac{t^3}{3} \right]_{0}^{x} = x^3 \]
\[ \int_{0}^{x} 3t \, dt = 3 \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{3x^2}{2} \]
\[ \int_{0}^{x} 1 \, dt = [t]_{0}^{x} = x \]

### Step 4: Combine the results
Now, we combine all the results from the integrations:
\[ \int_{0}^{x} (t + 1)^3 \, dt = \frac{x^4}{4} + x^3 + \frac{3x^2}{2} + x \]

### Step 5: Simplify the expression
To simplify, we can express everything in terms of a common denominator:
\[ = \frac{x^4}{4} + \frac{4x^3}{4} + \frac{6x^2}{4} + \frac{4x}{4} \]
\[ = \frac{x^4 + 4x^3 + 6x^2 + 4x}{4} \]

### Step 6: Factor the polynomial
We can factor the polynomial in the numerator:
\[ x^4 + 4x^3 + 6x^2 + 4x = x(x^3 + 4x^2 + 6x + 4) \]

### Step 7: Final expression
Thus, the final expression for the integral is:
\[ \int_{0}^{x} (t + 1)^3 \, dt = \frac{x(x^3 + 4x^2 + 6x + 4)}{4} \]

### Step 8: Consider the greatest integer function
Since the problem states that we need to evaluate this for \( [x] \), we will substitute \( [x] \) into our expression:
\[ \int_{0}^{[x]} (t + 1)^3 \, dt = \frac{[x]([x]^3 + 4[x]^2 + 6[x] + 4)}{4} \]

### Conclusion
The value of the integral \( \int_{0}^{[x]} (t + 1)^3 \, dt \) is:
\[ \frac{[x]([x]^3 + 4[x]^2 + 6[x] + 4)}{4} \]