question_answer

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

A) \[\frac{5}{6}\] sq. unit

B) \[\frac{6}{5}\] sq. unit

C) \[\frac{1}{6}\] sq. unit

D) 6 sq. unit

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 \[=\int_{0}^{1}{({{x}^{2}}+x)dx=\left[ \frac{{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2} \right]_{0}^{1}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}}\] sq. unit.