`int_(-1)^(2) sqrt(5x+6)dx`

Answer

To solve the definite integral \( \int_{-1}^{2} \sqrt{5x+6} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t^2 = 5x + 6 \). This means we can express \( x \) in terms of \( t \):
\[ x = \frac{t^2 - 6}{5} \]

### Step 2: Differentiate to find \( dx \)
Differentiating both sides with respect to \( t \):
\[ dx = \frac{2t}{5} dt \]

### Step 3: Change the limits of integration
Now we need to change the limits of integration from \( x \) to \( t \).

- For the lower limit when \( x = -1 \):
\[ t^2 = 5(-1) + 6 = 1 \implies t = 1 \quad (\text{since } t \geq 0) \]

- For the upper limit when \( x = 2 \):
\[ t^2 = 5(2) + 6 = 16 \implies t = 4 \quad (\text{since } t \geq 0) \]

### Step 4: Rewrite the integral
Now we can rewrite the integral in terms of \( t \):
\[ \int_{-1}^{2} \sqrt{5x + 6} \, dx = \int_{1}^{4} \sqrt{t^2} \cdot \frac{2t}{5} dt \]
Since \( \sqrt{t^2} = t \) for \( t \geq 0 \), we have:
\[ = \int_{1}^{4} t \cdot \frac{2t}{5} dt = \frac{2}{5} \int_{1}^{4} t^2 \, dt \]

### Step 5: Integrate
Now we can integrate \( t^2 \):
\[ \int t^2 \, dt = \frac{t^3}{3} \]
Thus,
\[ \frac{2}{5} \int_{1}^{4} t^2 \, dt = \frac{2}{5} \left[ \frac{t^3}{3} \right]_{1}^{4} \]

### Step 6: Evaluate the definite integral
Now we evaluate the limits:
\[ = \frac{2}{5} \left( \frac{4^3}{3} - \frac{1^3}{3} \right) = \frac{2}{5} \left( \frac{64}{3} - \frac{1}{3} \right) = \frac{2}{5} \left( \frac{63}{3} \right) = \frac{2 \cdot 63}{15} = \frac{126}{15} = \frac{42}{5} \]

### Final Answer
Thus, the value of the definite integral is:
\[ \int_{-1}^{2} \sqrt{5x + 6} \, dx = \frac{42}{5} \]

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