3.0 Area between the Curves

  Area of Bounded Regions

If we have two functions $f(x)$ and $g(x)$ such that $f(x) \leqslant g(x)\forall x \in \left( {a,b} \right)$, then the area bounded by the curves $y=f(x)$, $y=g(x)$ and the lines $x=a$, $x=b$ is given by $$A = \int\limits_a^b {[f(x) - g(x)]dx} $$

Question 4. Find the area bounded by ${y^2} = 4ax$ and ${x^2} = 4by$.

Solution: From given data, let $$f(x) = \sqrt {4ax} $$ and $$g(x) = \frac{{{x^2}}}{{4b}}$$

For the intersection point of both the curves,$${y^2} = 4ax \Rightarrow x = \frac{{{y^2}}}{{4a}}....(1)$$We have ${x^2} = 4by$, implies$$\begin{equation} \begin{aligned} {\left( {\frac{{{y^2}}}{{4a}}} \right)^2} = 4by \\ \Rightarrow y = 0\ and\ {y^3} = {4^3}{a^2}b \\ (or)\ y = 0\ and\ y = 4{a^{\frac{2}{3}}}{b^{\frac{1}{3}}...(2)} \\\end{aligned} \end{equation} $$Then $x$-coordinate of point of intersection is given by (2) and ${x^2} = 4by$. Let it be $k$.

Then the required area is given by $$\begin{equation} \begin{aligned} A = \int\limits_0^k {\left( {\sqrt {4ax} - \frac{{{x^2}}}{{4b}}} \right)dx} \\ \Rightarrow A = \left[ {\sqrt {4a} \left( {{x^{\frac{3}{2}}}} \right)\times\left( {\frac{2}{3}} \right) - \left( {\frac{1}{{4b}}} \right)\times\left( {\frac{{{x^3}}}{3}} \right)} \right]_0^k \\ \Rightarrow A = \frac{{16ab}}{3} \\\end{aligned} \end{equation} $$

Question 5. Find the area between the curves ${y^2} = 4ax$ and $y = mx$.

Solution: From given data,let $$\begin{equation} \begin{aligned} f(x) = \sqrt {4ax} \\ g(x) = mx \\\end{aligned} \end{equation} $$For the intersection point,

$$\begin{equation} \begin{aligned} f(x) = g(x) \\ \Rightarrow mx = \sqrt {4ax} \\ \Rightarrow {\left( {mx} \right)^2} = 4ax \\ \Rightarrow x({m^2}x - 4a) = 0 \\ \Rightarrow x = 0\ and\ x = \frac{{4a}}{{{m^2}}} \\\end{aligned} \end{equation} $$Let $c = \frac{{4a}}{{{m^2}}}$. Then the area bounded by the curves is given by $$\begin{equation} \begin{aligned} A = \int\limits_0^c {f(x) - g(x)dx} \\ \Rightarrow A = \int\limits_0^c {(\sqrt {4ax} - mx)dx} \\ \Rightarrow A = \sqrt {4a} \int\limits_0^c {{x^{\frac{1}{2}}}dx} - m\int\limits_0^c {xdx} \\ \Rightarrow A = \sqrt {4a} \times\left( {\frac{2}{3}} \right)[{x^{\frac{3}{2}}}]_0^c - m[\frac{{{x^2}}}{2}]_0^c \\\end{aligned} \end{equation} $$On solving, we get$$A = \frac{{8{a^2}}}{{3{m^3}}}$$

Note:

Consider the case when $f(x)$ and $g(x)$ intersect each other in the interval $[a,b]$. In this case, first we find out the point where the curves intersect from $f(x)=g(x)$ and the root be $c$ (Here, we considered only one intersection point to explain the phenomenon).

Thus, the required(shaded) area is $$\int\limits_a^c {\{ f(x) - g(x)\} dx} + \int\limits_c^b {\{ f(x) - g(x)\} dx} $$

Question 6. Find the area bounded by the curve $y = {x^3}$ , the $x$-axis and the ordinates $x=-2$ and $x=2$.

Solution: From the curve, it is clear that the curve $y = {x^3}$ intersect at the point $(0,0)$ in between $[-2,2]$.

Let $f(x) = {x^3}$ and $g(x) = 0$ (equation of the x-axis).

Then the required area,$$\begin{equation} \begin{aligned} A = \int\limits_{ - 2}^0 {\{ g(x) - f(x)\} dx} + \int\limits_0^2 {\{ f(x) - g(x)\} dx} \\ \Rightarrow A = \int\limits_{ - 2}^0 {\{ 0 - {x^3}\} dx} + \int\limits_0^2 {\{ {x^3} - 0\} dx} \\ \Rightarrow A = [\frac{{ - {x^4}}}{4}]_{ - 2}^0 + [\frac{{{x^4}}}{4}]_0^2 \\ \Rightarrow A = \left\{ {0 - \left( {\frac{{ - {{( - 2)}^4}}}{4}} \right)} \right\} + \left\{ {\frac{{{2^4}}}{4} - 0} \right\} \\ \Rightarrow A = 4 + 4 \\ \Rightarrow A = 8\ sq.units \\\end{aligned} \end{equation} $$

Question 7. Find the area of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$.

Solution: Using the symmetry of the curve, the area of ellipse is given by

$$\begin{equation} \begin{aligned} A = 4(areaOABO) \\ \Rightarrow A = 4\int\limits_0^a {ydx} \\\end{aligned} \end{equation} $$We have $$\begin{equation} \begin{aligned} \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \\ \Rightarrow \frac{{{y^2}}}{{{b^2}}} = 1 - \frac{{{x^2}}}{{{a^2}}} \\ \Rightarrow {y^2} = \frac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right) \\ \Rightarrow y = \pm \frac{b}{a}\left( {\sqrt {{a^2} - {x^2}} } \right) \\\end{aligned} \end{equation} $$But $y$ is positive above x-axis. Hence,$ \Rightarrow y = + \frac{b}{a}\left( {\sqrt {{a^2} - {x^2}} } \right)$.

Therefore,$$\begin{equation} \begin{aligned} \Rightarrow A = 4\int\limits_0^a {\left( {\frac{b}{a}\left( {\sqrt {{a^2} - {x^2}} } \right)} \right)dx} \\ \Rightarrow A = \frac{{4b}}{a}\int\limits_0^a {\left( {\sqrt {{a^2} - {x^2}} } \right)dx} \\ \Rightarrow A = \frac{{4b}}{a}\left\{ {\frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right\}_0^a \\ \Rightarrow A = \frac{{4b}}{a}\left[ {\left\{ {0 + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{a}{a}} \right\} - \left\{ {0 + 0} \right\}} \right] \\ \Rightarrow A = \frac{{4b}}{a}\left( {\frac{{{a^2}}}{2}{{\sin }^{ - 1}}1} \right) \\ \Rightarrow A = 2ab\left( {\frac{\pi }{2}} \right) \\ \Rightarrow A = \pi ab\ sq.units \\\end{aligned} \end{equation} $$ $\therefore $Area of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is $\pi ab\ sq.units$

Question 8. Prove that for any functions $f(x)$ and $g(x)$, integrable on the interval $(a,b)$, $$\left| {\int\limits_a^b {f(x)g(x)dx} } \right| \leqslant \sqrt {\left( {\int\limits_a^b {{f^2}(x)dx} } \right)\left( {\int\limits_a^b {{g^2}(x)dx} } \right)} $$

Solution: Consider a function $F(x) = {\left\{ {f(x) - \lambda g(x)} \right\}^2}$, where $\lambda $ is any real number. Since $F(x)$ is positive $\forall x \in (a,b)$ (i.e, the curve lies above the $x$-axis) the area under the curve is positive (from the above conventions followed),$$\begin{equation} \begin{aligned} \Rightarrow \int\limits_a^b {F(x)dx} \geqslant 0 \\ \Rightarrow \int\limits_a^b {{{\left\{ {f(x) - \lambda g(x)} \right\}}^2}dx} \geqslant 0 \\ \Rightarrow \int\limits_a^b {\left[ {{\lambda ^2}{{\left\{ {g(x)} \right\}}^2} + {{\left\{ {f(x)} \right\}}^2} - 2\lambda f(x)g(x)} \right]} \geqslant 0 \\ \Rightarrow {\lambda ^2}\left[ {\int\limits_a^b {{{\left\{ {g(x)} \right\}}^2}dx} } \right] - \lambda \left[ {\int\limits_a^b {2f(x)g(x)dx} } \right] + \left[ {\int\limits_a^b {{{\left\{ {f(x)} \right\}}^2}dx} } \right] \geqslant 0\ ...(1) \\\end{aligned} \end{equation} $$The above inequality (1), is a quadratic trinomial which is always positive, implies, the curve never touches x-axis. Hence, the discriminant is negative,$$\begin{equation} \begin{aligned} {\left\{ {\int\limits_a^b {2f(x)g(x)dx} } \right\}^2} - 4\left\{ {\int\limits_a^b {{{\left\{ {g(x)} \right\}}^2}dx} } \right\}\left\{ {\int\limits_a^b {{{\left\{ {f(x)} \right\}}^2}dx} } \right\} \leqslant 0 \\ \Rightarrow {\left\{ {\int\limits_a^b {f(x)g(x)dx} } \right\}^2} \leqslant \left\{ {\int\limits_a^b {{{\left\{ {g(x)} \right\}}^2}dx} } \right\}\left\{ {\int\limits_a^b {{{\left\{ {f(x)} \right\}}^2}dx} } \right\} \\ \Rightarrow \left| {\int\limits_a^b {f(x)g(x)dx} } \right| \leqslant \sqrt {\left\{ {\int\limits_a^b {{{\left\{ {g(x)} \right\}}^2}dx} } \right\}\left\{ {\int\limits_a^b {{{\left\{ {f(x)} \right\}}^2}dx} } \right\}} \\\end{aligned} \end{equation} $$Hence, proved.