Definite Integration Question 200
Question: The slope of the tangent to a curve $ y=f(x) $ at $ (x,f(x)) $ is $ 2x+1 $ . If the curve passes through the point (1, 2), then the area of the region bounded by the curve, the x-axis and the line $ x=1 $ is
Options:
A) $ \frac{5}{6} $ sq. unit
B) $ \frac{6}{5} $ sq. unit
C) $ \frac{1}{6} $ sq. unit
D) 6 sq. unit
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We, given, $ \frac{dy}{dx}=2x+1\Rightarrow y=x^{2}+x+k $ Since, the curve passes through the point (1, 2).
$ \therefore 2=1+1+k\Rightarrow k=0 $
$ \therefore $ The curve is $ y=x^{2}+x $ . So, the required area $ =\int_0^{1}{(x^{2}+x)dx=[ \frac{x^{3}}{3}+\frac{x^{2}}{2} ]_0^{1}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}} $ sq. unit.
BETA
