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Anettt [7]
3 years ago
15

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

Mathematics
1 answer:
RSB [31]3 years ago
6 0

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.

   

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