Definite Integration Question 418
Question: The area of the closed bounded by $ x=-1, $ $ y=0, $ $ y=x^{2}+x+1 $ , and the tangent to the curve $ y=x^{2}+x+1 $ at A(1,3) is
Options:
A) 4/3 sq. units
B) 7/3 sq. units
C) 7/6 sq. units
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ y=x^{2}+x+1={{( x+\frac{1}{2} )}^{2}}+\frac{3}{4} $
$ \Rightarrow y-\frac{3}{4}={{( x+\frac{1}{2} )}^{2}} $
This is a parabola with vertex at $ ( -\frac{1}{2},\frac{3}{4} ) $ and the curve is concave upwards.
$ y=x^{2}+x+1 $ or $ \frac{dy}{dx}=2x+1 $ or $ {{( \frac{dy}{dx} )} _{(1.3)}}=3 $
Equation of the tangent at $ A(1,3) $ is $ y=3x $ Required area = Area ABDMN $ - $ Area ONA Now, area ABDMN $ =\int\limits _{-1}^{1}{(x^{2}+x+1)dx} $
$ =2\int\limits_0^{1}{(x^{2}+1)=\frac{8}{3}} $ Area of ONA $ =\frac{1}{2}\times 1\times 3=\frac{3}{2}. $
$ \therefore $ Required area $ =\frac{8}{3}-\frac{3}{2} $
$ =\frac{16-9}{6}=\frac{7}{6} $ sq. units.
BETA
