Area as Definite Integral

Let f (x) be a continuous function in (a, b). Then the area bounded by the curve y = f (x),

x axis and lines x = a and x = b is given by the formula

$A=\left|\underset{a}{\overset{b}{\int }}f(x)dx\right|$,

provided f (x) $⩾$ 0 (or f(x) $⩽$ 0) $\forall$ x $\in$ (a , b)

It is sometimes convenient to use formula for area with respect to y i.e. regarding x as a function of y.

The area between x = f(y), y axis and the lines

y = c and y = d is given by

$A=\underset{c}{\overset{d}{\int }}f(y)dy$

If we have two functions f(x) and g(x) such that f(x) $⩽$ g(x) $\forall$ x $\in$ [a, b], then the area bounded by the curves y = f(x), y = g(x) and lines x = a,

x = b (a < b) is given by

$A=\underset{a}{\overset{b}{\int }}[g(x)-f(x)]dx$

(i) If curve lies completely above the x axis, then the area is positive but when it lies completely below x axis, then the area is negative, however we have the convention to consider the magnitude only.

(ii) If curve lies on both the sides of x axis i.e. above the x axis as well as below the x axis, then calculate both areas separately and add their modulus to get the total area.

In general if curve y = f(x) crosses the x axis n times when x varies from a to b, then the area between y = f(x), x axis and lines x = a and x = b is given by

$A=\left|A_1\right|+\left|A_2\right|+.........+\left|A_n\right|$

(iii) If the curve is symmetrical about x axis, or y axis, or both, then calculate the area of one symmetrical part and multiply it by the number of symmetrical parts to get the whole area.

Illustration 1: Find the area between the curves y = x² + x –2 and y = 2x, for which |x² + x –2| + | 2x | = |x² + 3x –2| is satisfied.

Solution: y = x² + x - 2 Þ y = 2x

|x² + x - 2| + |2x| = |x² + 3x - 2|

(x² + x - 2) and 2x have same sign

Thus required area

ar (PQR) + ar (ECD)

= $\underset{-1}{\overset{0}{\int }}$[2x - (x² + x - 2)] dx + $\underset{1}{\overset{2}{\int }}$[2x - (x² + x - 2)] dx

= ${\left[\frac{x^2}{2}-\frac{x^3}{3}+2x\right]}_{-1}^0+{\left[\frac{x^2}{2}-\frac{x^3}{3}+2x\right]}_1^2$

= $\frac{7}{6}+\left[\frac{10}{3}-\frac{13}{6}\right]=\frac{7}{6}+\frac{7}{6}=\frac{7}{3}$

Illustration 2 : Find out the area enclosed by circle |z| = 2, parabola y = x² + x + 1, the curve y = $\left[{\sin }^2\frac{x}{4}+\cos \frac{x}{4}\right]$ and x-axis (when [ . ] is the greatest integer function).

Solution: For x $\in$ [-2, 2]

$\Rightarrow$ 1 < sin² $\frac{x}{4}+\cos \frac{x}{4}$ < 2

$\therefore$ [sin² $\frac{x}{4}+\cos \frac{x}{4}$] = 1

Now we have to find out the area enclosed by the circle |z| = 2,

parabola (y - $\frac{3}{4}$) = ${\left(x+\frac{1}{2}\right)}^2$,

line y = 1and x-axis.

$\therefore$ Required area is shaded area in the figure.

Hence required area = $\sqrt{3}\times 1+(\sqrt{3}-1)+\underset{-1}{\overset{0}{\int }}(x^2+x+1).dx+2\underset{\sqrt{3}}{\overset{2}{\int }}\sqrt{4-x^2}$dx

= $\left(\frac{2\pi }{3}+\sqrt{3}-\frac{1}{6}\right)$ sq. units

Illustration 3: Let f(x) = Max. {sinx, cos x, $\frac{1}{2}$} then determine the area of the region bounded by the curves y = f(x), x-axis, y-axis and x = 2$\pi$.

Solution: $\because$. f(x) = Max { sin x, cos x, $\frac{1}{2}$}

interval value of f(x)

for 0 $⩽$ x < $\frac{\pi }{4}$, cos x

for $\frac{\pi }{4}$ $⩽$ x < $\frac{5\pi }{6}$, sin x

for $\frac{5\pi }{6}$$⩽$ x < $\frac{5\pi }{3}$, $\frac{1}{2}$

for $\frac{5\pi }{3}\le x\le 2\pi$ cos x

Hence required area

= $\underset{0}{\overset{\pi /4}{\int }}\cos xdx+\underset{\pi /4}{\overset{5\pi /6}{\int }}\sin xdx+\underset{5\pi /6}{\overset{5\pi /3}{\int }}\frac{1}{2}.dx+\underset{5\pi /3}{\overset{2\pi }{\int }}\cos xdx$

= $\left(\frac{5\pi }{12}+\sqrt{2}+\sqrt{3}\right)$ sq units.