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.

`( 27 )/4`

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`

=> `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)

=`0 – [( 81 )/4– 27 ] `

= `– [( 81 –108)/4]`

= `( 27 )/4`

The required area

=`( 0 )/4 + 0–[ (–3)^(4 )/4+(–3)3 ]`=`0 – [( 81 )/4– 27 ] `

= `– [( 81 –108)/4]`

= `( 27 )/4`

` ( 2 )/3. [8 – 1] = ( 14 )/3`

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`

= ` –∫_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`

© 2022 iPracticeMath | All Rights Reserved | Terms of Use.