Question

The area (in sq unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ and the ordinates $x=1$, $x=2$is

• A. $\frac{17}{12}$
• B. $\frac{11}{6}$
• C. $\frac{13}{3}$
• D. $1$

Answer 1

The correct option is A

$\frac{17}{12}$

Determining the area bounded by the curve.

Given, curves are parabola $y=x^2$, $y=x^3$ and the lines $x=1$ and $x=2$

Now the graph of the given curves is as shown.

Let the required area be denoted by $A$, therefore,

$$\begin{gathered} A={\int }_1^2\left(x^3-x^2\right)dx \\ ={{\left[\frac{x^{{\displaystyle 4}}}{{\displaystyle 4}}-\frac{x^{{\displaystyle 3}}}{{\displaystyle 3}}\right]}_1}^2 \\ =\left(\frac{2^{{\displaystyle 4}}}{{\displaystyle 4}}-\frac{2^{{\displaystyle 3}}}{{\displaystyle 3}}\right)-\left(\frac{1^{{\displaystyle 4}}}{{\displaystyle 4}}-\frac{1^{{\displaystyle 3}}}{{\displaystyle 3}}\right) \\ =\frac{16}{{\displaystyle 4}}-\frac{8}{{\displaystyle 3}}-\frac{1}{{\displaystyle 4}}+\frac{1}{{\displaystyle 3}} \\ =\frac{15}{{\displaystyle 4}}-\frac{7}{{\displaystyle 3}} \\ =\frac{\left(15\times 3\right)-\left(7\times 4\right)}{{\displaystyle 12}} \\ =\frac{45-28}{{\displaystyle 12}} \\ =\frac{17}{12} \end{gathered}$$

Therefore the area bounded by the given curve is equal to $\frac{17}{12}\text{square units}$

Hence, the correct answer is option (A).