Question

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

Answer 1

Answer:

Please check the explanation.

Step-by-step 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

$\int _{-2}^2x^2dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

   $=\left[\frac{x^{2+1}}{2+1}\right]^2_{-2}$

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

computing the boundaries

     $=\frac{16}{3}$

Thus,

$\int _{-2}^2x^2dx=\frac{16}{3}$

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

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

  $=\frac{16}{3}-16$

  $=-\frac{32}{3}$

  $=-10.67$ 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.