Area under the curve `y=sqrt(3x+4)` between x=0 and x=4 is

Options

A. `56/9` sq. units
B. `64/9` sq. units
C. 8 sq. units
D. `112/9` sq. units

Answer

To find the area under the curve \( y = \sqrt{3x + 4} \) between \( x = 0 \) and \( x = 4 \), we will use definite integration. Here’s the step-by-step solution:

### Step 1: Set up the integral
The area \( A \) under the curve from \( x = 0 \) to \( x = 4 \) can be expressed as:
\[ A = \int_{0}^{4} \sqrt{3x + 4} \, dx \]

### Step 2: Use substitution
To simplify the integration, we will use the substitution method. Let:
\[ t = 3x + 4 \]
Then, differentiate to find \( dx \):
\[ dt = 3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3} \]

### Step 3: Change the limits of integration
Now we need to change the limits of integration based on our substitution:
- When \( x = 0 \):
\[ t = 3(0) + 4 = 4 \]
- When \( x = 4 \):
\[ t = 3(4) + 4 = 16 \]

So, the new limits of integration will be from \( t = 4 \) to \( t = 16 \).

### Step 4: Substitute in the integral
Now substituting \( t \) and \( dx \) into the integral, we have:
\[ A = \int_{4}^{16} \sqrt{t} \cdot \frac{dt}{3} \]
This simplifies to:
\[ A = \frac{1}{3} \int_{4}^{16} t^{1/2} \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ \int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \]
Thus, we have:
\[ A = \frac{1}{3} \cdot \left[ \frac{2}{3} t^{3/2} \right]_{4}^{16} \]
This simplifies to:
\[ A = \frac{2}{9} \left[ t^{3/2} \right]_{4}^{16} \]

### Step 6: Evaluate the definite integral
Now we evaluate the limits:
\[ A = \frac{2}{9} \left( 16^{3/2} - 4^{3/2} \right) \]
Calculating \( 16^{3/2} \) and \( 4^{3/2} \):
\[ 16^{3/2} = (4^2)^{3/2} = 4^3 = 64 \]
\[ 4^{3/2} = (2^2)^{3/2} = 2^3 = 8 \]
Thus:
\[ A = \frac{2}{9} (64 - 8) = \frac{2}{9} \cdot 56 = \frac{112}{9} \]

### Final Answer
The area under the curve \( y = \sqrt{3x + 4} \) between \( x = 0 \) and \( x = 4 \) is:
\[ \frac{112}{9} \text{ square units} \]

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