The area (in sq. units) bounded between `y=3sinx and y =-4sin^(3)x` from `x=0` to `x=pi` is

Options

A. `4pi`
B. `34pi`
C. 4
D. `(34)/(3)`

Answer

To find the area bounded between the curves \( y = 3 \sin x \) and \( y = -4 \sin^3 x \) from \( x = 0 \) to \( x = \pi \), we will follow these steps:

### Step 1: Identify the curves
The two curves are:
1. \( y = 3 \sin x \)
2. \( y = -4 \sin^3 x \)

### Step 2: Find the points of intersection
To find the area between the curves, we first need to determine where they intersect. We set the two equations equal to each other:
\[ 3 \sin x = -4 \sin^3 x \]
Rearranging gives:
\[ 4 \sin^3 x + 3 \sin x = 0 \]
Factoring out \( \sin x \):
\[ \sin x (4 \sin^2 x + 3) = 0 \]
This gives us two cases:
1. \( \sin x = 0 \) which occurs at \( x = 0 \) and \( x = \pi \).
2. \( 4 \sin^2 x + 3 = 0 \) has no real solutions since \( \sin^2 x \) cannot be negative.

Thus, the points of intersection are \( x = 0 \) and \( x = \pi \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = \pi \) is given by:
\[ A = \int_{0}^{\pi} (3 \sin x - (-4 \sin^3 x)) \, dx \]
This simplifies to:
\[ A = \int_{0}^{\pi} (3 \sin x + 4 \sin^3 x) \, dx \]

### Step 4: Simplify the integral
We can factor out \( \sin x \):
\[ A = \int_{0}^{\pi} \sin x (3 + 4 \sin^2 x) \, dx \]

### Step 5: Use the identity for \( \sin^2 x \)
Recall that \( \sin^2 x = 1 - \cos^2 x \), so we can rewrite the integral:
\[ A = \int_{0}^{\pi} \sin x (3 + 4(1 - \cos^2 x)) \, dx \]
This becomes:
\[ A = \int_{0}^{\pi} \sin x (7 - 4 \cos^2 x) \, dx \]

### Step 6: Change of variables
Let \( t = \cos x \), then \( dt = -\sin x \, dx \). The limits change as follows:
- When \( x = 0 \), \( t = 1 \)
- When \( x = \pi \), \( t = -1 \)

Thus, the integral becomes:
\[ A = \int_{1}^{-1} (7 - 4t^2)(-dt) = \int_{-1}^{1} (7 - 4t^2) \, dt \]

### Step 7: Evaluate the integral
Now we evaluate:
\[ A = \int_{-1}^{1} (7 - 4t^2) \, dt = \left[ 7t - \frac{4t^3}{3} \right]_{-1}^{1} \]
Calculating this gives:
\[ = \left( 7(1) - \frac{4(1)^3}{3} \right) - \left( 7(-1) - \frac{4(-1)^3}{3} \right) \]
\[ = \left( 7 - \frac{4}{3} \right) - \left( -7 + \frac{4}{3} \right) \]
\[ = 7 - \frac{4}{3} + 7 - \frac{4}{3} = 14 - \frac{8}{3} = \frac{42}{3} - \frac{8}{3} = \frac{34}{3} \]

### Final Answer
Thus, the area bounded between the curves from \( x = 0 \) to \( x = \pi \) is:
\[ \boxed{\frac{34}{3}} \text{ square units} \]