The value of `int_(0)^(3) xsqrt(1+x)dx`, is

Options

A. `(9)/(2)`
B. `(27)/(4)`
C. `(126)/(15)`
D. `(116)/(15)`

Answer

To solve the integral \( I = \int_{0}^{3} x \sqrt{1+x} \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = 1 + x \). Then, differentiating both sides gives us:
\[ dx = dt \]
Also, we need to change the limits of integration. When \( x = 0 \), \( t = 1 + 0 = 1 \). When \( x = 3 \), \( t = 1 + 3 = 4 \).

### Step 2: Express \( x \) in terms of \( t \)
From our substitution, we have:
\[ x = t - 1 \]
Now we can rewrite the integral in terms of \( t \):
\[ I = \int_{1}^{4} (t - 1) \sqrt{t} \, dt \]

### Step 3: Expand the integrand
Now we can expand the integrand:
\[ I = \int_{1}^{4} (t \sqrt{t} - \sqrt{t}) \, dt = \int_{1}^{4} (t^{3/2} - t^{1/2}) \, dt \]

### Step 4: Integrate term by term
Now we can integrate each term separately:
\[ I = \int_{1}^{4} t^{3/2} \, dt - \int_{1}^{4} t^{1/2} \, dt \]

Calculating the first integral:
\[ \int t^{3/2} \, dt = \frac{t^{5/2}}{5/2} = \frac{2}{5} t^{5/2} \]
Calculating the second integral:
\[ \int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \]

### Step 5: Evaluate the integrals from 1 to 4
Now we evaluate both integrals from 1 to 4:
\[ I = \left[ \frac{2}{5} t^{5/2} \right]_{1}^{4} - \left[ \frac{2}{3} t^{3/2} \right]_{1}^{4} \]

Calculating the first part:
\[ \frac{2}{5} \left( 4^{5/2} - 1^{5/2} \right) = \frac{2}{5} \left( 32 - 1 \right) = \frac{2}{5} \cdot 31 = \frac{62}{5} \]

Calculating the second part:
\[ \frac{2}{3} \left( 4^{3/2} - 1^{3/2} \right) = \frac{2}{3} \left( 8 - 1 \right) = \frac{2}{3} \cdot 7 = \frac{14}{3} \]

### Step 6: Combine the results
Now we combine the results:
\[ I = \frac{62}{5} - \frac{14}{3} \]

### Step 7: Find a common denominator
The common denominator for 5 and 3 is 15:
\[ I = \frac{62 \cdot 3}{15} - \frac{14 \cdot 5}{15} = \frac{186}{15} - \frac{70}{15} = \frac{116}{15} \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{\frac{116}{15}} \]