Applications of Integrals

This chapter applies the principles of definite integration to calculate the area of bounded regions. The fundamental concept is that the area under a curve can be approximated by summing the areas of a large number of thin vertical or horizontal rectangular strips. As the width of these strips approaches zero, the sum becomes a definite integral, giving the exact area. By the end of this chapter, you will be expected to visualize a region described by given equations, set up the correct integral to represent its area, and then evaluate it.

The main topics are finding the area bounded by a single curve and the coordinate axes, and the more common problem of finding the area enclosed between two or more curves. This involves finding the points of intersection of the curves, which define the limits of integration. You will learn to decide whether to integrate with respect to x (using vertical strips) or with respect to y (using horizontal strips) based on which approach simplifies the calculation.

The primary difficulty in this chapter is not usually the integration itself, but the initial setup. A correct and reasonably accurate sketch of the curves is essential to identify the required region and the correct limits. This chapter is a direct application of Definite Integrals and also heavily depends on your knowledge of Functions and Conic Sections for curve sketching.

Formulas for Applications of Integrals

Area under the curve y = f(x): $A = \int_{a}^{b} y \, dx = \int_{a}^{b} f(x) \, dx$ Area bounded by y=f(x), the x-axis, and the lines x=a and x=b.
Area bounded by the curve x = g(y): $A = \int_{c}^{d} x \, dy = \int_{c}^{d} g(y) \, dy$ Area bounded by x=g(y), the y-axis, and the lines y=c and y=d.
Area between two curves y = f(x) and y = g(x): $A = \int_{a}^{b} [f(x) - g(x)] \, dx$ Where f(x) ≥ g(x) for x in [a, b]. f(x) is the upper curve.
Area between two curves x = f(y) and x = g(y): $A = \int_{c}^{d} [f(y) - g(y)] \, dy$ Where f(y) ≥ g(y) for y in [c, d]. f(y) is the right-hand curve.
Area of a circle x² + y² = r²: $A = 4 \int_{0}^{r} \sqrt{r^2 - x^2} \, dx = \pi r^2$ Calculated by finding the area in the first quadrant and multiplying by 4.
Area of an ellipse x²/a² + y²/b² = 1: $A = 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^2 - x^2} \, dx = \pi ab$ The final result is πab square units.

One of these questions, worked through

JEE Advanced 2013

The area enclosed by the curves $y = \sin x + \cos x$ and $y = |\cos x − \sin x|$ over the interval [0, π/2] is $(1) 4(√2 − 1) (2) 2√2(√2 − 1) (3) 2(√2 + 1) (4) 2√2(√2 + 1)$

Applications of Integrals The area enclosed by the curves

y = \sin x + \cos x

and

y = |\cos x − \sin x|

over the interval [0, π/2] is

(1) 4(√2 − 1) (2) 2√2(√2 − 1) (3) 2(√2 + 1) (4) 2√2(√2 + 1)

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$y_1 = \sin x + \cos x, \qquad y_2 = |\cos x - \sin x|$ $x \in \left[0, \tfrac{\pi}{2}\right]$ $\text{Remove the modulus, find upper/lower, then } A = \int_0^{\pi/2} (\text{upper} - \text{lower})\,dx$ $|\cos x - \sin x| = \begin{cases} \cos x - \sin x & x \in [0,\tfrac{\pi}{4}] \\[2pt] \sin x - \cos x & x \in [\tfrac{\pi}{4},\tfrac{\pi}{2}] \end{cases}$ $y_1 - y_2 = 2\sin x \geq 0 \quad \left(0 \leq x \leq \tfrac{\pi}{4}\right)$ $y_1 - y_2 = 2\cos x \geq 0 \quad \left(\tfrac{\pi}{4} \leq x \leq \tfrac{\pi}{2}\right)$ $\Rightarrow y_1 \text{ is on top throughout}$ $A = \int_0^{\pi/4} 2\sin x\,dx + \int_{\pi/4}^{\pi/2} 2\cos x\,dx$ $\int_0^{\pi/4} 2\sin x\,dx = [-2\cos x]_0^{\pi/4} = -\tfrac{2}{\sqrt{2}} + 2 = 2\left(1 - \tfrac{1}{\sqrt{2}}\right)$ $\int_{\pi/4}^{\pi/2} 2\cos x\,dx = [2\sin x]_{\pi/4}^{\pi/2} = 2 - \tfrac{2}{\sqrt{2}} = 2\left(1 - \tfrac{1}{\sqrt{2}}\right)$ $A = 4\left(1 - \tfrac{1}{\sqrt{2}}\right) = 4 - 2\sqrt{2} = 2\sqrt{2}(\sqrt{2} - 1)$ $\Rightarrow \text{option (2)}$ $A = 2\sqrt{2}(\sqrt{2} - 1) \text{ square units}$

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Past-paper questions on Applications of Integrals

JEE Advanced 22 questions