Find the area between the curves y2−y+x=0 and y2=x+3.
Solution:
Let x2(y)=y−y2, x1(y)=y2−3.
Let's draw the graphs of these functions.
We have two points of intersection.
y−y2=y2−3⇒2y2−y−3=0.
y1=1.5⇒x1=−0.75.
y2=−1⇒x2=−2.
Find the area S.
S=∫y2y1(x2(y)−x1(y))dy.
S=∫−11.5((y−y2)−(y2−3))dy=∫−11.5(−2y2+y+3)dy.
S=(−2)∫−11.5y2dy+∫−11.5ydy+3∫−11.5dy= (−2)3y3∣∣−11.5+2y2∣∣−11.5+3y∣−11.5=24125.
Answer:
S=24125
Comments
Leave a comment