Application Of Integrals, Revision Notes: CBSE Class 12-humanities APPLIED MATHEMATICS, Applied Mathematics

Application of Integrals

Area of the region bounded by the curve y = f(x), x-axis, and the lines x = a and x = b (b > a) is given by $A = \int_{a}^{b} y ⅆ x$ or $A = \int_{a}^{b} f (x) ⅆ x$
The area of the region bounded by the curve x = g(y), y-axis, and the lines y = c and y = d is given by $A = \int_{a}^{b} x ⅆ y$ or $A = \int_{c}^{d} g (y) ⅆ x$
If a line y = mx + p intersects a curve y = f(x) at x = a and x = b, (b > a), then the area (A) of region bounded by the curve y = f(x) and the line y = mx + p is

$A = \int_{a}^{b} (y_{1} - y_{2}) ⅆ x, where y_{1} = m x + p and y_{2} = f (x)$

$A = \int_{a}^{b} [(m x + p) - f (x)] d x$

If a line y = mx + p intersects a curve x = g(y) at y = c and y = d ,(d > c), then the area (A) of region bounded by the curve x = g(y) and the line y = mx + p is

$A = \int_{c y}^{d} (x_{1} - x_{2}) ⅆ y, where x_{1} = \frac{y - p}{m} and x_{2} = g (y) A = \int_{c}^{d} [(\frac{y - p}{m}) - g (y)] ⅆ y$

Example 1: Find the area of the region in the first and third quadrant enclosed by the x-axis and the line $y = \sqrt{3 x}$ , and the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Solution: The given equations are

$y = \sqrt{3 x}$ ... (1)

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ... (2)

Substituting $y = \sqrt{3 x}$ in equation (2), we obtain

$\frac{x^{2}}{a^{2}} + \frac{3 x^{2}}{b^{2}} = 1 \Rightarrow x^{2} (b^{2} + 3 a^{2}) = a^{2} b^{2} \Rightarrow x = \pm \frac{a b}{\sqrt{b^{2} + 3 a^{2}}} ∴ y = \pm \frac{\sqrt{3} a b}{\sqrt{b^{2} + 3 a^{2}}}$

Hence, the line meets the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at C $(\frac{a b}{\sqrt{b^{2} + 3 a^{2}}}, \frac{\sqrt{3} a b}{\sqrt{b^{2} + 3 a^{2}}})$ and D $(\frac{- a b}{\sqrt{b^{2} + 3 a^{2}}}, \frac{- \sqrt{3} a b}{\sqrt{b^{2} + 3 a^{2}}})$ in the first and third quadrant respectively.

In the figure, CM ⊥ XX′

Now, area OCMO = $\int_{0}^{\frac{a b}{\sqrt{b^{2} + 3 a^{2}}}} \sqrt{3} x ⅆ x = \frac{\sqrt{3}}{2} {[x^{2}]}_{0}^{\frac{a b}{\sqrt{b^{2} + 3 a^{2}}}} = \frac{\sqrt{3} a^{2} b^{2}}{2 (b^{2} + 3 a^{2})}$

Area ACMA

$= \int_{\frac{a b}{\sqrt{b^{2} + 3 a^{2}}}}^{a} \frac{b}{a} \sqrt{a^{2} - x^{2}} ⅆ x = \frac{b}{a} {[\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a}]}_{\frac{a b}{\sqrt{b^{2} + 3 a^{2}}}}^{a} = \frac{b}{a} [\frac{a}{2} \times 0 + \frac{a^{2}}{2} \times \sin^{- 1} 1 - (\frac{a b}{2 \sqrt{b^{2} + 3 a^{2}}} \times \frac{\sqrt{3} a^{2}}{\sqrt{b^{2} + 3 a^{2}}} + \frac{a^{2}}{2} \sin^{- 1} \frac{b}{\sqrt{b^{2} + 3 a^{2}}})] = \frac{π}{4} a b - \frac{\sqrt{3} a^{2} b^{2}}{2 (b^{2} + 3 a^{2})} - \frac{a b}{2} \sin^{- 1} \frac{b}{\sqrt{b^{2} + 3 a^{2}}}$

The area of the region enclosed bet…

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