Texas Instruments TI-82 User Manual - Page 214

Finding the Area between Curves

Page 214 highlights

Finding the Area between Curves Find the area of the region bounded by: f(x) = 300 x à (x2 + 625) g(x) = 3 cos .1 x x = 75 Procedure 1. Press z. Select the default MODE settings. Press o and turn off all functions. Press y [STAT PLOT] and turn off all stat plots. 2. Press p. Set the viewing WINDOW. Xmin = 0 Xmax = 100 Xscl = 10 Ymin = M5 Ymax = 10 Yscl = 1 3. Press o. Enter the upper and lower functions. Y1=300Xà(X2+625) Y2=3cos .1X 4. Press y ãCALCä and select intersection. The graph is displayed. Select First curve, Second curve, and Guess for the intersection at the left of the display. The solution is displayed, and the value of X at the intersection, which is the lower limit of the integral, is stored in Ans and X. 5. Press y ãDRAWä and use Shade( to see the area graphically. Shade(Y2,Y1,1,Ans,75) 6. Press y ãQUITä to return to the Home screen. Enter the expression to evaluate the integral for the shaded region. fnInt(Y1-Y2,X,Ans,75) The area is 325.839962. 14-20 Applications

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14-20
Applications
Finding the Area between Curves
Find the area of the region bounded by:
f(x) = 300 x
à
(x
2
+ 625)
g(x) = 3 cos
.
1 x
x = 75
Procedure
1.
Press
z
. Select the default
MODE
settings. Press
o
and turn off all
functions. Press
y
[
STAT PLOT
] and turn off all stat plots.
2.
Press
p
. Set the viewing
WINDOW
.
Xmin = 0
Ymin =
M
5
Xmax = 100
Ymax = 10
Xscl = 10
Yscl = 1
3.
Press
o
. Enter the upper and lower functions.
Y
1
=300X
à
(X
2
+625)
Y
2
=3cos .1X
4.
Press
y
ã
CALC
ä
and select
intersection
. The graph is displayed. Select
First curve
,
Second curve
, and
Guess
for the intersection at the left of
the display. The solution is displayed, and the value of
X
at the
intersection, which is the lower limit of the integral, is stored in
Ans
and
X
.
5.
Press
y
ã
DRAW
ä
and use
Shade(
to see the area graphically.
Shade(Y
2
,Y
1
,1,Ans,75)
6.
Press
y
ã
QUIT
ä
to return to the Home screen. Enter the expression to
evaluate the integral for the shaded region.
fnInt(Y
1
–Y
2
,X,Ans,75)
The area is
325.839962
.