Application of Multiple Integrals

The most important application of Integrals involves finding areas bounded by a curve and x-axis. It includes findings solutions to the problems of work and energy.
In the next section, we present some examples of finding areas under the curve with the help of definite integrals.


Multiple Integral Examples


Example 1:

 `( 27 )/4`

Explanation:

We first find the points where the curve cuts the x-axis. We put `y = 0`


`x^3 + 3x^2 = 0`

=> `x^2(x + 3) = 0`

=> `x = 0 and x = -3`

Therefore, the curve crosses the x-axis at (-3 , 0) and (0 , 0) … So we apply the integration limits from -3 to 0.

The required area =`∫_(-3)^0(x^(3 )+ 3x^2 ) dx` = ` [ x^(4 )/4]+ x3 ` (Applying the limits)

The required area

=`( 0 )/4 + 0–[ (–3)^(4 )/4+(–3)3 ]`

=`0 – [( 81 )/4– 27 ] `

= `– [( 81 –108)/4]`

= `( 27 )/4`

Example 2:

 ` ( 2 )/3. [8 – 1] = ( 14 )/3`

Explanation:

The part of the curve above the x-axis is `y = √(4-x)`

We put ` 4 – x = t ` ` – dx = dt ` => ` dx = – dt `

So For the value of `x = 0 => 4 – 0 = t => t = 4 `

And for `x = 3 => 4 – 3 = t => t = 1`


=>The Required Area Under the curve is given by


`∫_0^3√(4-x) dx = ∫_4^1√t (–dt)`

= ` –∫_4^1√t (dt)`

= ` t^(3/2 )/(3/2)`

( Applying the limits of `t = 4 to t = 1 )`

Area under the curve =`( 2 )/3. [ (4)^(3/2)– (1)^(3/2)] =( 2 )/3. [8 – 1] = ( 14 )/3`





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