The area of the region bounded by the x-axis and the curves defined by $y = t an x$ , $- \frac{π}{3} \leq x \leq \frac{π}{3}$ and $y = co t x$ , $\frac{π}{6} \leq x \leq \frac{3 π}{2}$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — $l o g 2$

Detailed Solution

The curves meet where $t an x = co t x$ , i.e. at $x = \frac{π}{4}$ (height 1). The closed region with the x-axis lies under $y = t an x$ from $0$ to $\frac{π}{4}$ and under $y = co t x$ from $\frac{π}{4}$ to $\frac{π}{2}$ .

$\int_{0}^{π /4} t an x d x = [l n sec x]_{0}^{π /4} = l n 2 = \frac{1}{2} l n 2$ .

$\int_{π /4}^{π /2} co t x d x = [l n s in x]_{π /4}^{π /2} = 0 - l n \frac{1}{2} = \frac{1}{2} l n 2$ . Total area $= l n 2$ (log here is the natural log).

✓ Check

co t x = t an (\frac{π}{2} - x)

, so the region is symmetric about

x = \frac{π}{4}

and the two parts must be equal.

⚠ Common trap

(B)

\frac{1}{2} l o g 2

is only the part under

y = t an x

; the part under

y = co t x

is forgotten.

Quick tip

$\int_{0}^{π /4} tan x d x = \int_{π /4}^{π /2} cot x d x = \frac{1}{2} ln 2$ , so the total is $ln 2$

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Application of Definite Integration — Preparation

This question tests your understanding of Application of Definite Integration, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

This topic uses definite integrals to find the area of regions in the plane. It covers the area under a curve and above the x-axis, the area between a curve and the y-axis, the area between two curves, and the areas of standard regions bounded by circles, parabolas, ellipses and straight lines. Always start with a rough sketch of the region and find where the curves intersect, since those points become the limits of integration. For the area between two curves, integrate (upper curve - lower curve) between their intersection points. If part of a curve lies below the x-axis, take the absolute value of that part separately rather than integrating straight across, or the positive and negative areas will cancel. Useful standard results include the area of the ellipse x^2/a^2 + y^2/b^2 = 1, which is pi ab. Practise the area under the curve questions below, each with a sketch and a step-by-step solution.

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