Find the area between the curve `(x ^(2))/( a ^(2)) + (y ^(2))/(b ^(2)) =1 and ` the x-axis between `x =0 and x =a.` Draw rough sketch of the curve also.

Answer

To find the area between the curve \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (which represents an ellipse) and the x-axis from \(x = 0\) to \(x = a\), we can follow these steps:

### Step 1: Understand the Equation of the Ellipse
The given equation of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
This represents an ellipse centered at the origin with semi-major axis \(b\) along the y-axis and semi-minor axis \(a\) along the x-axis.

### Step 2: Find the y-coordinate in terms of x
To find the area under the curve, we need to express \(y\) in terms of \(x\):
\[ \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} \]
\[ y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right) \]
Taking the square root (since we are interested in the upper half of the ellipse):
\[ y = b \sqrt{1 - \frac{x^2}{a^2}} \]

### Step 3: Set Up the Integral for Area
The area \(A\) between the curve and the x-axis from \(x = 0\) to \(x = a\) can be expressed as:
\[ A = \int_{0}^{a} y \, dx \]
Substituting the expression for \(y\):
\[ A = \int_{0}^{a} b \sqrt{1 - \frac{x^2}{a^2}} \, dx \]

### Step 4: Factor out the constant \(b\)
\[ A = b \int_{0}^{a} \sqrt{1 - \frac{x^2}{a^2}} \, dx \]

### Step 5: Use Substitution to Solve the Integral
Let \(x = a \sin(\theta)\), then \(dx = a \cos(\theta) d\theta\). The limits change as follows:
- When \(x = 0\), \(\theta = 0\)
- When \(x = a\), \(\theta = \frac{\pi}{2}\)

Substituting into the integral:
\[ A = b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin^2(\theta)} \cdot a \cos(\theta) \, d\theta \]
Using the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\):
\[ A = b \int_{0}^{\frac{\pi}{2}} a \cos^2(\theta) \, d\theta \]

### Step 6: Solve the Integral
The integral of \(\cos^2(\theta)\) can be solved using the identity:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Thus,
\[ A = b \cdot a \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \]
Calculating the integral:
\[ = \frac{ab}{2} \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{2}} \]
Evaluating the limits:
\[ = \frac{ab}{2} \left[ \frac{\pi}{2} + 0 - (0 + 0) \right] = \frac{ab\pi}{4} \]

### Final Result
The area between the curve and the x-axis from \(x = 0\) to \(x = a\) is:
\[ A = \frac{ab\pi}{4} \]