MHT CET · Maths · Application of Derivatives
The slope of the tangent to a curve \(y=\mathrm{f}(x)\) at \((x, \mathrm{f}(x))\) is \(2 x+1\). If the curve passes through the point \((1,2)\), then the area (in sq. units), bounded by the curve, the \(\mathrm{X}\)-axis and the line \(x=1\), is
- A \(\frac{3}{2}\)
- B \(\frac{4}{3}\)
- C \(\frac{5}{6}\)
- D \(\frac{1}{12}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{6}\)
Step-by-step Solution
Detailed explanation
According to the given condition, \(\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x+1\) Integrating w.r.t. \(x\), we get \(y=x^2+x+\mathrm{c}\) As curve passes through \((1,2)\), we get \(2=(1)^2+1+c \Rightarrow c=0\)
\(\therefore \quad\) The equation of the curve is \(y=x^2+x\).
\(\therefore\) Required area
\(\begin{aligned} & =\int_0^1\left(x^2+x\right) \mathrm{d} x \\ & =\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^1 \\ & =\frac{1}{3}+\frac{1}{2} \\ & =\frac{5}{6} \text { sq. units }\end{aligned}\)
\(\therefore \quad\) The equation of the curve is \(y=x^2+x\).
\(\therefore\) Required area
\(\begin{aligned} & =\int_0^1\left(x^2+x\right) \mathrm{d} x \\ & =\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^1 \\ & =\frac{1}{3}+\frac{1}{2} \\ & =\frac{5}{6} \text { sq. units }\end{aligned}\)
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