Question

How do you determine the area under a curve in calculus using integrals or the limit definition of integrals?

Answer 1

Please check the explanation.

Explanation:

Let us consider

`y = f(x)`

To find the area under the curve
`y = f(x)` between
`x = a` and
`x = b`, all we need is to integrate
`y = f(x)` between the limits of
`a` and
`b`.

For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:

`A=\int _a^b|f\left(x\right)|dx`

=
`\int _(-2)^2\left|x^2-4\right|dx`

`\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx`

`=\int _(-2)^2x^2dx-\int _(-2)^24dx`

solving

`\int _(-2)^2x^2dx`

`\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=(x^(a+1))/(a+1),\:\quad \:a\\e -1`

`=\left[(x^(2+1))/(2+1)\right]^2_(-2)`

`=\left[(x^3)/(3)\right]^2_(-2)`

computing the boundaries

`=(16)/(3)`

Thus,

`\int _(-2)^2x^2dx=(16)/(3)`

similarly solving

`\int _(-2)^24dx`

`\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax`

`=\left[4x\right]^2_(-2)`

computing the boundaries

`=16`

Thus,

`\int _(-2)^24dx=16`

Therefore, the expression becomes

`A=\int _a^b|f\left(x\right)|dx=\int _(-2)^2x^2dx-\int _(-2)^24dx`

`=(16)/(3)-16`

`=-(32)/(3)`

`=-10.67` square units

Thus, the area under a curve is -10.67 square units

The area is negative because it is below the x-axis. Please check the attached figure.