Definite Integrals (VSAQs)

Maths-2B | 7. Definite Integrals – VSAQs:
Welcome to VSAQs in Chapter 7: Definite Integrals. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : Evaluate ∫₁⁵ dx/√2x-1

$$I = \int_{1}^{5} \frac{dx}{\sqrt{2x-1}} = \left[ \frac{1}{\sqrt{2}} \cdot 2\sqrt{2x-1} \right]_{1}^{5}$$

$$= \left[ \sqrt{2} \cdot \sqrt{2x-1} \right]_{1}^{5}$$

$$= \left[ \sqrt{2(5)-1} – \sqrt{2(1)-1} \right]$$

$$= \sqrt{9} – \sqrt{1}$$

$$= 3 – 1$$

$$= 2$$

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VSAQ-2 : Evaluate ∫₀³ xdx/√x²+16

$$I = \int_{0}^{3} \frac{x dx}{\sqrt{x^2+16}} = \frac{1}{2} \int_{0}^{3} \frac{2x dx}{\sqrt{x^2+16}}$$

$$= \left[ \sqrt{x^2+16} \right]_{0}^{3} = \left[ \sqrt{9+16} – \sqrt{0^2+16} \right]$$

$$= \sqrt{25} – \sqrt{16}$$

$$= 5 – 4$$

$$= 1$$

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VSAQ-3 : Evaluate ∫₂³ 2xdx/1+x²

$$\int_{2}^{3} \frac{2x dx}{1+x^2} = \left[ \log |1+x^2| \right]_{2}^{3}$$

$$= \log |1 + 3^2| – \log |1 + 2^2|$$

$$= \log 10 – \log 5$$

$$= \log \frac{10}{5}$$

$$= \log 2$$

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VSAQ-4 : Evaluate ∫₀^π √2+2cosθ dθ

$$I = \int_{0}^{\pi} \sqrt{2+2\cos\theta} d\theta = \int_{0}^{\pi} \sqrt{2(1 + \cos\theta)} d\theta$$

$$= \int_{0}^{\pi} \sqrt{2 \cdot 2\cos^2\left(\frac{\theta}{2}\right)}d\theta = \int_{0}^{\pi} 2\cos\left(\frac{\theta}{2}\right) d\theta$$

$$= 2 \cdot 2\left[\sin\left(\frac{\theta}{2}\right)\right]_{0}^{\pi} = 4\left(\sin\left(\frac{\pi}{2}\right) – \sin(0)\right)$$

$$= 4(1 – 0)$$

$$= 4$$

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VSAQ-5 : Evaluate ∫₀¹ x²/1+x² dx

$$\int_{0}^{1} \frac{x^2}{1+x^2} dx = \int_{0}^{1} \left(1 – \frac{1}{1+x^2}\right) dx$$

$$= \int_{0}^{1} 1 dx – \int_{0}^{1} \frac{1}{1+x^2} dx$$

$$= [x]{0}^{1} – [\tan^{-1}x]{0}^{1}$$

$$= (1 – 0) – (\tan^{-1}1 – \tan^{-1}0)$$

$$= 1 – \frac{\pi}{4}$$

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VSAQ-6 : Evaluate ∫₀⁴ x²/1+x dx

$$I = \int_{0}^{4} (x^2 – 1) + \frac{1}{1 + x} dx$$

$$= \int_{0}^{4} (x^2 – 1)dx + \int_{0}^{4} \frac{dx}{1 + x}$$

$$= \left[\frac{x^3}{3} – x\right]{0}^{4} + \left[\log(1 + x)\right]{0}^{4}$$

$$= \left[\frac{4^3}{3} – 4\right] + \log(5) – \log(1)$$

$$= \frac{64}{3} – 4 + \log(5)$$

$$= \frac{64 – 12}{3} + \log(5)$$

$$= \frac{52}{3} + \log(5)$$

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VSAQ-7 : Evaluate ∫₀^a dx/x²+a²

$$\int_{0}^{a} \frac{dx}{x^2 + a^2} = \left[\frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)\right]_{0}^{a}$$

$$= \frac{1}{a}\left[\tan^{-1}\left(\frac{a}{a}\right) – \tan^{-1}\left(\frac{0}{a}\right)\right]$$

$$= \frac{1}{a}\left[\tan^{-1}(1) – \tan^{-1}(0)\right]$$

$$= \frac{1}{a}\left[\frac{\pi}{4} – 0\right]$$

$$= \frac{\pi}{4a}$$

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VSAQ-8 : Evaluate ∫₀^a √a²-x² dx

Using the formula:

$$\int \sqrt{a^2 – x^2} dx = \frac{x}{2} \sqrt{a^2 – x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$$

Evaluating from 0 to a:

$$\int_{0}^{a} \sqrt{a^2 – x^2} dx = \left[\frac{x}{2} \sqrt{a^2 – x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right)\right]_{0}^{a}$$

$$= \left(0 + \frac{a^2}{2} \sin^{-1}\left(\frac{a}{a}\right)\right) – \left(\frac{0}{2} \sqrt{a^2 – 0^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{0}{a}\right)\right)$$

$$= \frac{a^2}{2} \sin^{-1}(1)$$

Since $$\sin^{-1}(1) = \frac{\pi}{2}$$

$$= \frac{a^2}{2} \left(\frac{\pi}{2}\right) = \frac{\pi a^2}{4}$$

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VSAQ-9 : Evaluate ∫₀² |1 – x| dx

From the definition of the modulus function, we have:

$$|1 – x| = 1 – x \text{ for } 1 – x \geq 0$$

$$\Rightarrow x \leq 1$$

$$|1 – x| = -(1 – x) = x – 1 \text{ for } 1 – x < 0$$

$$\Rightarrow x > 1$$

$$\int_{0}^{2} |1 – x| dx = \int_{0}^{1} (1 – x)dx + \int_{1}^{2} (x – 1)dx$$

$$= \left[x – \frac{x^2}{2}\right]{0}^{1} + \left[\frac{x^2}{2} – x\right]{1}^{2}$$

$$= \left(1 – \frac{1}{2}\right) + \left(\frac{4}{2} – 2\right) – \left(\frac{1}{2} – 1\right)$$

$$= \frac{1}{2} + (2 – 2) – \left(\frac{1}{2} – 1\right)$$

$$= \frac{1}{2} + 0 – \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1$$

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VSAQ-10 : Evaluate ∫₀⁴ |2 – x| dx

$$|2 – x| = 2 – x \text{ when } 2 – x \geq 0$$

$$\Rightarrow x \leq 2$$

$$|2 – x| = -(2 – x) \text{ when } 2 – x < 0$$

$$\Rightarrow x > 2$$

$$\int_{0}^{4} |2 – x| dx = \int_{0}^{2} (2 – x) dx + \int_{2}^{4} (x – 2) dx$$

$$= \left[2x – \frac{x^2}{2}\right]{0}^{2} + \left[\frac{x^2}{2} – 2x\right]{2}^{4}$$

$$= \left(4 – \frac{4}{2}\right) + \left(\frac{16}{2} – 8\right) – \left(\frac{4}{2} – 4\right)$$

$$= 2 + (8 – 8) – (2 – 4)$$

$$= 2 + 0 – (-2)$$

$$= 2 + 2$$

$$= 4$$

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VSAQ-11 : Evaluate ∫₀^π∕2 sin⁴ xdx

When n is even, the formula for the integral of $$\sin^n x$$ from 0 to $$\frac{\pi}{2}$$ is given

$$\int_{0}^{\frac{\pi}{2}} \sin^n x dx = \frac{(n – 1)(n – 3) \ldots (3)}{n(n – 2) \ldots (4)} \cdot \frac{1}{2} \cdot \frac{\pi}{2}$$

For $$\int_{0}^{\frac{\pi}{2}} \sin^4 x dx$$

$$n = 4$$

$$\int_{0}^{\frac{\pi}{2}} \sin^4 x dx = \frac{(3)(1)}{(4)(2)} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$$

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VSAQ-12 : Evaluate ∫₀^π∕2 cos¹¹ x dx

When n is odd

$$\int_{0}^{\frac{\pi}{2}} \cos^n x dx = \frac{(n – 1)(n – 3) \ldots}{n(n – 2) \ldots 2 \cdot 3} \cdot 1$$

For $$\int_{0}^{\frac{\pi}{2}} \cos^{11} x dx$$

$$n = 11$$

$$\int_{0}^{\frac{\pi}{2}} \cos^{11} x dx = \frac{(10)(8)(6)(4)(2)}{(11)(9)(7)(5)(3)} \cdot 1$$

$$= \frac{256}{693}$$

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VSAQ-13 : Evaluate ∫₀^π∕2 sin⁵ xcos⁴  xdx

$$\int_{0}^{\frac{\pi}{2}} \sin^5 x \cos^4 x dx = \left[\frac{(4)(2)}{(9)(7)(5)(3)}\right] \cdot \frac{\pi}{2} = \frac{8}{315}$$

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VSAQ-14 : Evaluate ∫₀^π∕2 sin⁶ xcos⁴ xdx

$$\int_{0}^{\frac{\pi}{2}} \sin^6 x \cos^4 x dx = \left[\frac{(5)(3)(1)}{(10)(8)(6)(4)(2)}\right] \pi/2 = \frac{3\pi}{512}$$

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VSAQ-15 : Evaluate ∫₀^π sin³ x.cos³ x dx

$$I = \int_{0}^{\pi} \sin^3(\pi – x) \cdot \cos^3(\pi – x) dx$$

$$= -\int_{0}^{\pi} \sin^3 x \cdot \cos^3 x dx = -1$$

$$\Rightarrow 2I = 0$$

$$\Rightarrow I = 0$$

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VSAQ-16 : Evaluate ∫_-π∕2^π∕2 sin² xcos⁴ x dx

Let $$f(x) = \sin^2x \cos^4x$$

$$f(-x) = \sin^2(-x)\cos^4(-x) = (\sin x)^2(\cos x)^4 = \sin^2x\cos^4x = f(x)$$

Since f(x) is an even function,

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^4 xdx = 2 \int_{0}^{\frac{\pi}{2}} \sin^2x\cos^4 xdx$$

$$= 2 \left[\frac{(1)}{(6)(4)(2)} \pi\right] = \frac{\pi}{16}$$

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VSAQ-17 : Evaluate ∫₀^2π sin⁴ xcos⁶ xdx

Let $$f(x) = \sin^4x\cos^6x$$

Given $$f(2\pi – x) = f(\pi – x) = f(x)$$

$$\int_{0}^{2\pi} \sin^4 x \cos^6 x dx = 2\int_{0}^{\pi} \sin^4 x \cos^6 x dx$$

$$= 2(2) \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^6 x dx$$

$$= 4 \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^6 x dx$$

$$= 4 \left[\frac{(3)(1)}{(10)(8)(6)(4)(2)} \pi/2\right] = \frac{3\pi}{128}$$

$$= 4 \left( \frac{3}{10} \cdot \frac{1}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \right) = \frac{3\pi}{128}$$

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VSAQ-18 : Find ∫₀^2π sin² xcos⁴ xdx

Let $$f(x) = \sin^2 x \cos^4 x$$

Given $$f(2\pi – x) = f(\pi – x) = f(x)$$

$$\int_{0}^{2\pi} \sin^2 x \cos^4 x dx = 2\int_{0}^{\pi} \sin^2 x \cos^4 x dx$$

$$= 2 \cdot 2 \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^4 x dx$$

$$= 4 \left[ \frac{(1)}{(6)(4)(2)} \pi/2 \right] = \frac{\pi}{8}$$

$$= 4 \left( \frac{1}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \right) = \frac{\pi}{8}$$