- Here we need to find the the area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b
- we can think of area under the curve as composed of large number of very thin vertical strips.
- Consider an arbitrary strip of height y and width dx, then dA (area of the elementary strip)= ydx, where, y = f(x).

- Total area between x=a and x=b can be obtained by adding up the elementary areas of thin strips across the region
- A= $\int_{a}^{b} f(x) \; dx$

- Similary The area A of the region bounded by the curve x = g (y), y-axis and the lines y = a,y = b is given as

- A= $\int_{a}^{b} g(y) \; dy$

A= |$\int_{a}^{b} f(x) \; dx$ |

If the position of the graph under consideration is below x axis for some part and above x axis for some part.then we need to calculate the integration of both the parts separately and then just add the positive value

A= $\int_{a}^{c} f(x) \; dx$ + |$\int_{c}^{b} f(x) \; dx|$

Lets take few example

(1) Find the area of the region bounded by the curve $y = x^2$ and the line y = 4.

(a)First we need to find the points where the curve and the line intersect. To do this, set the equation of the curve equal to the equation of the line and solve it

So, here $4=x^2$ or $x=\pm 2$, So point of intersection are (-2.4) and (2,4)

(b)we also see that that Area around y axis is symmetrical, so we can find on one side and double it

(c) Consider a vertical strip dx, The length of the strip will be Y coordinate of line - Y coordinate of curve

So Area of bounded region

$A=2\int_{0}^{2} (4-x^2) \; dx= \frac {32}{3}$

$A= \int_{a}^{b} [f(x) -g(x)] \; dx$ , where, $f (x)\geq g (x)$ in [a, b]

(ii) If $f (x) \geq g (x)$ in [a, c] and $f (x) \leq g (x)$ in [c, b], a < c < b, then

$A= \int_{a}^{c} [f(x) -g(x)] \; dx + \int_{c}^{b} [g(x) -f(x)] \; dx$

2. Find the points where the curve and the line intersect. To do this, set the equation of the curve equal to the equation of the line and solve for \(x\) (or \(y\) if it's more suitable).

3. After plotting, determine which function is the top function and which one is the bottom for the interval you are interested in. You will subtract the bottom function from the top one to find the height of the thin strip (or differential element) that you'll use in integration.

4.To find the area, integrate as follows:

\[ \text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx \]

Where:

- \(f(x)\) is the top function

- \(g(x)\) is the bottom function

- \(a\) and \(b\) are the x-values of the points of intersection found in step 2.

5. Compute the integral to get the area.

Find the area of the region bounded by the curve \(y = x^2\) and the line \(y = 2x\).

1. Plot them. You'll see they intersect in two places.

2. Find points of intersection:

Setting \(x^2 = 2x\), we get \(x^2 - 2x = 0 \implies x(x-2) = 0\). So, the points are \(x = 0\) and \(x = 2\).

3. From the graph, between \(x = 0\) and \(x = 2\), \(2x\) is above \(x^2\).

4. Setup the integral:

\[ \text{Area} = \int_{0}^{2} [2x - x^2] \, dx \]

\[ = \left[ x^2 - \frac{x^3}{3} \right]_0^2 \]

\[ = \left[4 - \frac{8}{3}\right] \]

\[ = \frac{4}{3} \]

So, the area between the curve and the line in this interval is \( \frac{4}{3} \) square units.

Find the area between the curves \(y = \sqrt{x}\) and \(y = x^2\).

1. Plot them. You'll see they intersect in two places.

Set \(x^2 = \sqrt{x}\).

\[x^4 = x\]

\[x^4 - x = 0\]

\[x(x^3 - 1) = 0\]

This gives \(x = 0\) and \(x = 1\) as the points of intersection.

2. In this interval, \(y = \sqrt{x}\) is above \(y = x^2\).

3. Calculate the area

\[ \text{Area} = \int_{0}^{1} (\sqrt{x} - x^2) \, dx \]

\[ = \left[ \frac{2x^{3/2}}{3} - \frac{x^3}{3} \right]_0^1 \]

\[ = \frac{2}{3} - \frac{1}{3} \]

\[ = \frac{1}{3} \]

Find the area bounded by the curve y = cos x between x = 0 and $x = 2 \pi$.

First plot the curve

A= $\int_{0}^{\pi/2} cos x \; dx$ + |$\int_{\pi/2}^{3\pi/2} cos x \; dx$| + $\int_{3\pi/2}^{2\pi} cos x \; dx$

A=4

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