Mathematics 1

Yüklə 2,11 Mb.
ölçüsü2,11 Mb.
  1   2   3   4
MES-2 4th week

MES 2-4th week

Definite Integrals

  • As you remember the second way of think of integrals is as of the area under a curve (or to be more precise between the curve and x –axes) which leads to the idea of a definite integral, one with limits.
  • Area=.
  • It is easy, if the area
  • is rectangle

    If f(x) is given by the

    following function:

  • What is ?
  • Solution:
  • 192+106+192=136


Curves above and below the x-axis

  • If the curve is below the line the integral will be a negative number. To get the area (which must be positive) we must change the sign.
  • Hence if f(x) ≥ 0 :
  • = area between f(x) and
  • the x-axis.

  • If f(x) ≤ 0 :
  • =“-” area between f(x) and
  • the x-axis.

  • If a function crosses the x-axis
  • we have to split the region/

    integral into parts that are all

    above and all below and work out the areas for each part separately.


Curves above and below the x-axis-example1.49

  • What is the area bounded by the graph of y = (x−1)(x−2) and the coordinate axes?
  • Solution:
  • y = (x−1)(x−2)=x2-3x+2
  • y changes sign at x=1
  • and x=2

    Area A:


    Area B:=-(-+2)=(-+-4)==

    Total area enclosed: +=1


Area between two graphs

  • Sometimes we wish to find the area enclosed between two graphs. We can do this by subtracting the area of a smaller region from the area of a large region.
  • To determine the areas we need to determine the relevant interval for x to integrate over. To do this we must determine where the graphs intersect.

Area between two graphs - example1.51

  • What is the area bounded by the graphs of y = x2 and y = x3, and the lines x = 0 and x = 2?
  • Solution:

    In the Area A )

    Yüklə 2,11 Mb.

    Dostları ilə paylaş:
  1   2   3   4

Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur © 2024
rəhbərliyinə müraciət

    Ana səhifə