The area (in sq. units) bounded between the curves `y=e^(x)cos x and y=e^(x)sin x` from `x=(pi)/(4)` to `x=(pi)/(2)` is

Options

A. `(epi)/(2)`
B. `(e^((pi)/(4)))/(sqrt2)`
C. `(e^(lambda))/(sqrt2)`
D. `(e^(lambda))/(2)`

Answer

To find the area bounded between the curves \( y = e^x \cos x \) and \( y = e^x \sin x \) from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Identify the curves and the area formula
The area \( A \) between two curves \( y_1 \) and \( y_2 \) from \( x = a \) to \( x = b \) is given by:
\[ A = \int_a^b (y_2 - y_1) \, dx \]
In our case, we have:
- \( y_1 = e^x \cos x \)
- \( y_2 = e^x \sin x \)

### Step 2: Set up the integral
The area can be expressed as:
\[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (e^x \sin x - e^x \cos x) \, dx \]
This simplifies to:
\[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e^x (\sin x - \cos x) \, dx \]

### Step 3: Factor out \( e^x \)
We can factor out \( e^x \):
\[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e^x (\sin x - \cos x) \, dx \]

### Step 4: Use integration by parts
To evaluate the integral, we can use integration by parts. We set:
- \( u = \sin x - \cos x \) (thus \( du = (\cos x + \sin x) \, dx \))
- \( dv = e^x \, dx \) (thus \( v = e^x \))

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int e^x (\sin x - \cos x) \, dx = e^x (\sin x - \cos x) - \int e^x (\cos x + \sin x) \, dx \]

### Step 5: Evaluate the integral
Now we need to evaluate the integral \( \int e^x (\cos x + \sin x) \, dx \). This can be solved similarly using integration by parts again or recognizing it as a standard integral.

After evaluating, we find:
\[ \int e^x (\sin x - \cos x) \, dx = e^x (\sin x - \cos x) - e^x (\sin x + \cos x) + C \]

### Step 6: Apply limits
Now we apply the limits from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \):
\[ A = \left[ e^x (\sin x - \cos x - \sin x - \cos x) \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]

Calculating this gives:
\[ A = \left[ e^{\frac{\pi}{2}} (0) \right] - \left[ e^{\frac{\pi}{4}} \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right) \right] = e^{\frac{\pi}{2}} - e^{\frac{\pi}{4}} \cdot 0 = e^{\frac{\pi}{2}} \]

### Final Result
Thus, the area bounded between the curves from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \) is:
\[ A = e^{\frac{\pi}{2}} - e^{\frac{\pi}{4}} \cdot 0 = e^{\frac{\pi}{2}} \]