SATHEE: Area Question 38

Area Question 38

Question 38

$Missing or unrecognized delimiter for \left$ , where $0 \leq x \leq 1$ .

Determine the area of the region bounded by the curves $y = f (x), X$ -axis, $x = 0$ and $x = 1$ .

$(1997, 5$ M)

Show Answer

Solution:

We can draw the graph of $y = x^{2}, y = (1 - x^{2})$ and $y = 2 x (1 - x)$ in following figure

Now, to get the point of intersection of $y = x^{2}$ and $y = 2 x (1 - x)$ , we get

$\begin{aligned} x^{2} & = 2 x (1 - x) \Rightarrow & 3 x^{2} & = 2 x \Rightarrow & x (3 x - 2) & = 0 \Rightarrow & x & = 0, 2 / 3 \end{aligned}$

Similarly, we can find the coordinate of the points of intersection of

$y = (1 - x^{2}) and y = 2 x (1 - x) are x = 1 / 3 and x = 1$

From the figure, it is clear that,

$\begin{array}{cc} (1 - x)^{2}, & if 0 \leq x \leq 1 / 3 f (x) = 2 x (1 - x), & if 1 / 3 \leq x \leq 2 / 3 x^{2}, & if 2 / 3 \leq x \leq 1 \end{array}$

$∴$ The required area

$\begin{aligned} A & = \int_{0}^{1} f (x) d x & = \int_{0}^{1 / 3} (1 - x)^{2} d x + \int_{1 / 3}^{2 / 3} 2 x (1 - x) d x + \int_{2 / 3}^{1} x^{2} d x & = - \frac{1}{3} (1 - x)^{3} + x^{2} - {\frac{2 x^{3}}{3}}^{2 / 3} + \frac{1}{3} x^{3} & = - \frac{1}{3} {\frac{2}{3}}^{3} + \frac{1}{3} + {\frac{2}{3}}^{2} - \frac{2}{3} {\frac{2}{3}}^{3} - {\frac{1}{3}}^{2} + \frac{2}{3} {\frac{1}{3}}^{3} & + \frac{1}{3} (1) - \frac{1}{3} \frac{2^{3}}{3} \end{aligned}$

$= \frac{19}{81} + \frac{13}{81} + \frac{19}{81} = \frac{17}{27} sq unit$

Area Question 38

Question 38

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Area Question 38

Question 38

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu