Definite Integration Question 419
Question: The area of the closed figure bounded $ y=\frac{x^{2}}{2}-2x+2 $ and the tangents to it at (1, 1/2) and (4, 2) is
Options:
A) 9/8 sq. units
B) 3/8 sq. units
C) 3/2 sq. units
D) 9/4 sq. units
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ y=\frac{x^{2}}{2}-2x+2=\frac{{{(x-2)}^{2}}}{2}, $
$ \frac{dy}{dx}=x-2, $
$ {{( \frac{dy}{dx} )} _{x=1}}=-1, $
$ {{( \frac{dy}{dx} )} _{x=4}}=2, $
Thus, tangent at (1, 1/2) is $ y-1/2=-1(x-1) $ or $ 2x+2y-3=0 $ Tangent at $ (4,2) $ is $ y-2=2(x-4) $ or $ 2x-y-6=0 $
Hence, $ A=\int\limits_1^{5/2}{( \frac{x^{2}}{2}-2x+2-\frac{3-2x}{2} )dx} $
$ +\int\limits _{5/2}^{4}{( \frac{x^{2}}{2}-2x+2-(2x-6) )dx} $
$ =\int\limits_1^{4}{( \frac{x^{2}}{2}-2x+2 )dx-\int\limits_1^{5/2}{( \frac{3-2x}{2} )dx}} $
$ -\int\limits _{5/2}^{4}{(2x-6)dx} $
$ ={{( \frac{x^{3}}{6}-x^{2}+2x )}_1}^{4}-\frac{1}{2}{{(3x-x^{2})}_1}^{5/2}-( x^{2}-6x ) _{5/2}^{4} $ $ =( \frac{63}{6}-15+6 )-\frac{1}{2}( 3\times \frac{3}{2}-( \frac{25}{4}-1 ) )-( ( 16-\frac{25}{4} )-6( 4-\frac{5}{2} ) ) $ $ =\frac{3}{2}-\frac{1}{2}( \frac{9}{2}-\frac{21}{4} )-( \frac{39}{4}-6( \frac{3}{2} ) ) $ $ =\frac{9}{8}sq.units $
BETA
