10 Most SAQ’s of Definite Integrals Chapter in Inter 2nd Year Maths-2B (TS/AP)

4 Marks

SAQ-1 : Evaluate ∫₀^π∕2 a sin⁡ x+b cos ⁡x/sin⁡ x+cos ⁡x  dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx$$

Given:

$$I = \int_{0}^{\frac{\pi}{2}} \frac{a \sin x + b \cos x}{\sin x + \cos x} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{a \sin(\frac{\pi}{2} – x) + b \cos(\frac{\pi}{2} – x)}{\sin(\frac{\pi}{2} – x) + \cos(\frac{\pi}{2} – x)} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{a \cos x + b \sin x}{\cos x + \sin x} dx$$

Now adding them together:

$$I + I = \int_{0}^{\frac{\pi}{2}} \frac{a(\sin x + \cos x) + b(\cos x + sin x)}{\cos x + sin x} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{(a + b)(\cos x + \sin x)}{\cos x + \sin x} dx$$

$$= \int_{0}^{\frac{\pi}{2}} (a + b)dx$$

$$\Rightarrow 2I = (a + b) \left[ x \right]_{0}^{\frac{\pi}{2}}$$

$$\Rightarrow 2I = (a + b)\frac{\pi}{2}$$

$$\Rightarrow I = \frac{(a + b)\pi}{4}$$

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SAQ-2 : Evaluate ∫₀^π∕2 cos^5∕2 x/sin^5∕2+cos^5∕2 x dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a – x)dx$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{5/2} x}{\sin^{5/2} x + \cos^{5/2} x} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{\cos^{5/2}(\frac{\pi}{2} – x)}{\sin^{5/2}(\frac{\pi}{2} – x) + \cos^{5/2}(\frac{\pi}{2} – x)} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{\sin^{5/2} x}{\sin^{5/2} x + \cos^{5/2} x} dx$$

Now adding (1) & (2), we get:

$$I + I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{5/2} x}{\sin^{5/2} x + \cos^{5/2} x} dx + \int_{0}^{\frac{\pi}{2}} \frac{\sin^{5/2} x}{\sin^{5/2} x + \cos^{5/2} x} dx$$

$$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{5/2} x + \sin^{5/2} x}{\sin^{5/2} x + \cos^{5/2} x} dx$$

$$= \int_{0}^{\frac{\pi}{2}} 1 dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} – 0 = \frac{\pi}{2}$$

$$2I = \frac{\pi}{2}$$

$$\Rightarrow I = \frac{\pi}{4}$$

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SAQ-3 : Evaluate ∫₀^π∕2 sin⁵ x/sin⁵ x + cos⁵ x dx

We know $$\int_{0}^{a} f(x)dx = \int_{0}^{a} f(a – x)dx$$

$$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\sin^5 x + \cos^5 x} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{\cos^5(\frac{\pi}{2} – x)}{\sin^5(\frac{\pi}{2} – x) + \cos^5(\frac{\pi}{2} – x)} dx$$

$$= \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\sin^5 x + \cos^5 x} dx$$

Adding (1) and (2), we get:

$$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x + \cos^5 x}{\sin^5 x + \cos^5 x} dx$$

$$= \int_{0}^{\frac{\pi}{2}} 1 dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} – 0 = \frac{\pi}{2}$$

$$2I = \frac{\pi}{2}$$

$$\Rightarrow I = \frac{\pi}{4}$$

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SAQ-4 : Evaluate ∫_π∕6^π∕3 √sinx/√sinx + √cosx dx

We know $$\int_{a}^{b} f(x)dx = \int_{a}^{b} f(a + b – x)dx$$

Given:

$$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$

$$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin(\frac{\pi}{6} + \frac{\pi}{3} – x)}}{\sqrt{\sin(\frac{\pi}{6} + \frac{\pi}{3} – x)} + \sqrt{\cos(\frac{\pi}{6} + \frac{\pi}{3} – x)}} dx$$

$$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin(\frac{\pi}{2} – x)}}{\sqrt{\sin(\frac{\pi}{2} – x)} + \sqrt{\cos(\frac{\pi}{2} – x)}} dx$$

$$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$$

$$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$$

Adding (1) and (2):

$$I + I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$

$$\Rightarrow 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$$

$$= \left[ x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} – \frac{\pi}{6} = \frac{\pi}{6}$$

$$\Rightarrow 2I = \frac{\pi}{6}$$

$$\Rightarrow I = \frac{\pi}{12}$$

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SAQ-5 : Evaluate ∫₀^π∕2 dx/4 + 5 cos⁡ x

Given the substitution $$\tan \frac{x}{2} = t$$ then $$\cos x = \frac{1 – t^2}{1 + t^2}$$ and $$dx = \frac{2dt}{1 + t^2}$$

Also, when x = 0 t = 0 and when $$x = \frac{\pi}{2} t = 1$$

$$I = \int_{0}^{1} \frac{2dt}{(1 + t^2)} \cdot \frac{1}{4 + 5\left(\frac{1 – t^2}{1 + t^2}\right)}$$

$$= \int_{0}^{1} \frac{2dt}{(1 + t^2)} \cdot \frac{1}{4(1 + t^2) + 5(1 – t^2)}$$

$$= \int_{0}^{1} \frac{2dt}{1 + t^2} \cdot \frac{1}{9 – t^2}$$

$$= 2 \int_{0}^{1} \frac{dt}{9 – t^2}$$

$$= 2 \cdot \frac{1}{3} \log\left|\frac{3 + t}{3 – t}\right| \Bigg|_{0}^{1}$$

$$= \frac{2}{3} \left[\log\left|\frac{3 + 1}{3 – 1}\right| – \log\left|\frac{3 + 0}{3 – 0}\right|\right]$$

$$= \frac{2}{3} \log\left|\frac{4}{2}\right|$$

$$= \frac{2}{3} \log 2$$

$$= \frac{1}{3} \log 2$$

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SAQ-6 : Evaluate ∫₀⁴ (16-x²)^5∕2 dx

Given the substitution $$x = 4 \sin \theta$$

$$\Rightarrow dx = 4 \cos \theta d\theta$$

Also, when x = 0

$$\Rightarrow 4 \sin \theta = 0$$

$$\Rightarrow \sin \theta = 0$$

$$\Rightarrow \theta = 0$$

And when x = 4

$$\Rightarrow 4 \sin \theta = 4$$

$$\Rightarrow \sin \theta = 1$$

$$\Rightarrow \theta = \frac{\pi}{2}$$

Then the integral I becomes:

$$I = \int_{0}^{4} (16 – 16 \sin^2 \theta)^{5/2} \cdot 4 \cos \theta d\theta = 4 \int_{0}^{\frac{\pi}{2}} 64(1 – \sin^2 \theta)^{5/2} \cdot \cos \theta d\theta$$

$$= 4 \int_{0}^{\frac{\pi}{2}} 64 (\cos^2 \theta)^{5/2} \cdot \cos \theta d\theta$$

$$= 256 \int_{0}^{\frac{\pi}{2}} \cos^6 \theta d\theta$$

$$= 256 \cdot \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = 256 \cdot \frac{15}{48} \cdot \frac{\pi}{2} = 256 \cdot \frac{5}{16} \cdot \frac{\pi}{2}$$

$$= 16 \cdot 5 \cdot \frac{\pi}{2} = 80 \cdot \frac{\pi}{2} = 40\pi$$

$$I = 40\pi$$

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SAQ-7 : Find ∫_-a^a x²(a²-x²)^3∕2 dx

Given $$f(x) = x^2(a^2 – x^2)^{3/2}$$

Since $$f(-x) = (-x)^2(a^2 – (-x)^2)^{3/2} = x^2(a^2 – x^2)^{3/2} = f(x)$$

f(x) is an even function.

Hence $$\int_{-a}^{a} x^2(a^2 – x^2)^{3/2} dx = 2\int_{0}^{a} x^2(a^2 – x^2)^{3/2} dx$$

Let $$x = a \sin \theta$$ then $$dx = a \cos \theta d\theta$$

Now, for $$x = 0 a \sin \theta = 0 \sin \theta = 0 \theta = 0$$ and $$x = a a \sin \theta = a \sin \theta = 1 \theta = \frac{\pi}{2}$$

$$2\int_{0}^{a} x^2(a^2 – x^2)^{3/2} dx = 2\int_{0}^{\frac{\pi}{2}} a^2 \sin^2 \theta (a^2 – a^2 \sin^2 \theta)^{3/2} a \cos \theta d\theta$$

$$= 2a^6 \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \cos^4 \theta d\theta$$

$$= 2a^6 \frac{(3)(1)}{(6)(4)(2)} \cdot \frac{\pi}{2}$$

$$= 2a^6 \frac{3}{48} \cdot \frac{\pi}{2}$$

$$= \frac{a^6 \pi}{16}$$

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SAQ-8 : Find the area enclosed by the curves y=x² and y=x³

Solving $$y = x^2$$ and $$y = x^3$$ we have:

$$x^2 = x^3$$

$$\Rightarrow x^3 – x^2 = 0$$

$$\Rightarrow x^2(x – 1) = 0$$

$$\Rightarrow x = 0 \text{ or } x = 1$$

The upper boundary curve is $$y = x^3$$

The lower boundary curve is $$y = x^2$$

The required area A is given by:

$$A = \int_{0}^{1} [x^3 – x^2]dx = \left[\frac{x^4}{4} – \frac{x^3}{3}\right]_{0}^{1} = \frac{1}{4} – \frac{1}{3} = \frac{4 – 3}{12} = \frac{1}{12}$$

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SAQ-9 : Find the area of the region bounded by the parabola y²=4x and x²=4y

The given curves are $$y^2 = 4x$$ and $$x^2 = 4y$$

Solving (1) and (2) we get $$x^2 = 4y$$

$$\Rightarrow x^4 = 16y^2 = 16(4x) = 64x$$

$$\Rightarrow x^3 = 64$$

$$\Rightarrow x = 4$$

The upper boundary curve is $$y^2 = 4x \Rightarrow y = 2\sqrt{x}$$

The lower boundary curve is $$x^2 = 4y \Rightarrow y = x^2/4$$

$$A = \int_{0}^{4} (2\sqrt{x} – x^2/4)dx = \left[ \frac{4}{3}x^{3/2} – \frac{x^3}{12} \right]_{0}^{4} = \frac{32}{3} – \frac{64}{12} = \frac{16}{3} \text{ sq.units}$$

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SAQ-10 : Find the area of the region bounded by y²=4ax and x²=4by

Given curves are $$y^2 = 4ax$$ and $$x^2 = 4by$$

Solving (1) and (2), we get $$x^2 = 4by$$ which leads to $$x^4 = (16b^2)y^2 = 16b^2(4ax) = 64ab^2x$$

$$\Rightarrow x^3 = 64ab^2$$

$$\Rightarrow x = 0 or x^3 = 64ab^2$$

$$\Rightarrow x = 4(ab^2)^{1/3}$$

The upper boundary curve is $$y^2 = 4ax \Rightarrow y = 2\sqrt{ax}$$

The lower boundary curve is $$x^2 = 4by \Rightarrow y = \frac{x^2}{4b}$$

The area enclosed between the curves is:

$$A = \int_{0}^{4(ab^2)^{1/3}} \left(2\sqrt{ax} – \frac{x^2}{4b}\right)dx$$

$$= \left[ \frac{2\sqrt{a} \cdot 2}{3} x^{3/2} – \frac{1}{4b} \cdot \frac{x^3}{3} \right]_{0}^{4(ab^2)^{1/3}}$$

$$= \frac{4\sqrt{a}}{3} \left[4(ab^2)^{1/3}\right]^{3/2} – \frac{1}{12b} \left[4(ab^2)^{1/3}\right]^3$$

$$= \frac{4\sqrt{a}}{3} \cdot 2\sqrt{a} \cdot (b^2)^{1/2} – \frac{1}{12b} \cdot 64ab^2$$

$$= \frac{8a\sqrt{b}}{3} – \frac{16ab^2}{3b}$$

$$= \frac{8ab^{1/2}}{3} – \frac{16ab}{3}$$

$$= \frac{16ab}{3} \left(\frac{1}{2}b^{-1/2} – 1\right)$$

$$A = \frac{16ab}{3} \text{ sq.units}$$