# 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.

## Find the area bounded by the curve y = x^3 + 3x^2 and the x-axis

( 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

## Find the area between the x-axis and the curve y2 = 4 – x in the first quadrant from x = 0 to x = 3.

 ( 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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