I -44 x-2 d xii -33 x-1 d xiii -112 x-1 d xiv -222 x-3 d xv - 02 - 2-3 x-2 d xvi - 033 x-1 d xvii -66 x-2 d xviii -22 x-1 d xix - 12 x-3 d xx - 0 - - 2 cos 2 x d xxi - 02 -sin x d xxii - - 4 - - 4 sin x d xxiii - 28 x-5 d xxiv - - 2 - - 2 sin x-cos x d x

(i) #8747; 44x+2 #160;dxWe #160;know #160;that, #160;x+2= x+2 #160;, #160; 4 #8804;x #8804; 2x+2, #160; #160; #160; 2 #60;x #8804;4 ...

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Byju's Answer

Standard XII

Mathematics

Properties of Inequalities

i ∫-44 x+2 d ...

Question

(i) $\int_{- 4}^{4} |x + 2| d x$

(ii) $\int_{- 3}^{3} |x + 1| d x$

(iii) $\int_{- 1}^{1} |2 x + 1| d x$

(iv) $\int_{- 2}^{2} |2 x + 3| d x$

(v) $\int_{0}^{2} |x^{2} - 3 x + 2| d x$

(vi) $\int_{0}^{3} |3 x - 1| d x$

(vii) $\int_{- 6}^{6} |x + 2| d x$

(viii) $\int_{- 2}^{2} |x + 1| d x$

(ix) $\int_{1}^{2} |x - 3| d x$

(x) $\int_{0}^{π / 2} |\cos 2 x| d x$

(xi) $\int_{0}^{2 π} |\sin x| d x$

(xii) $\int_{- π / 4}^{π / 4} |\sin x| d x$

(xiii) $\int_{2}^{8} |x - 5| d x$

(xiv) $\int_{- π / 2}^{π / 2} \{\sin |x| + \cos |x|\} d x$

(xv) $\int_{0}^{4} |x - 1| d x$

(xvi) $\int_{1}^{4} \{|x - 1| + |x - 2| + |x - 4|\} d x$

(xvii) $\int_{- 5}^{0} f (x) d x, where f (x) = |x| + |x + 2| + |x + 5|$

(xviii) $\int_{0}^{4} (|x| + |x - 2| + |x - 4|) d x$

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Solution

(i)
$\int_{- 4}^{4} |x + 2| d x We know that, |x + 2| = \{\begin{cases} - (x + 2), - 4 \leq x \leq - 2 \\ x + 2, - 2 < x \leq 4 \end{cases} ∴ I = \int_{- 4}^{4} |x + 2| d x \Rightarrow I = \int_{- 4}^{- 2} - (x + 2) d x + \int_{- 2}^{4} (x + 2) d x \Rightarrow I = {[- \frac{x^{2}}{2} - 2 x]}_{- 4}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{4} \Rightarrow I = - 2 + 4 - 8 - 8 + 8 + 8 - 2 + 4 \Rightarrow I = 20$

(ii)
$I = \int_{- 3}^{3} |x + 1| d x We know that, |x + 1| = \{\begin{cases} - (x + 1), - 3 \leq x \leq - 1 \\ x + 1, - 1 < x \leq 3 \end{cases} ∴ I = \int_{- 3}^{- 1} - (x + 1) d x + \int_{- 1}^{3} [x + 1] d x \Rightarrow I = {[- \frac{{(x + 1)}^{2}}{2}]}_{- 3}^{- 1} + {[\frac{{(x + 1)}^{2}}{2}]}_{- 1}^{3} \Rightarrow I = 0 + 2 + 8 - 0 \Rightarrow I = 10$

(iii)
$\int_{- 1}^{1} |2 x + 1| d x We know that, |2 x + 1| = \{\begin{cases} - (2 x + 1), - 1 \leq x \leq - \frac{1}{2} \\ (2 x + 1), - \frac{1}{2} < x \leq 1 \end{cases} ∴ I = \int_{- 1}^{\frac{- 1}{2}} - (2 x + 1) d x + \int_{- \frac{1}{2}}^{1} (2 x + 1) d x \Rightarrow I = - {[x^{2} + x]}_{- 1}^{\frac{- 1}{2}} + {[x^{2} + x]}_{- \frac{1}{2}}^{1} \Rightarrow I = - \frac{1}{4} + \frac{1}{2} + 1 - 1 + 1 + 1 - \frac{1}{4} + \frac{1}{2} \Rightarrow I = \frac{5}{2}$

(iv)
$\int_{- 2}^{2} |2 x + 3| d x We know that, |2 x + 3| = \{\begin{cases} - (2 x + 3), - 2 \leq x \leq - \frac{3}{2} \\ (2 x + 3), - \frac{3}{2} < x \leq 2 \end{cases} ∴ I = \int_{- 2}^{\frac{- 3}{2}} - (2 x + 3) d x + \int_{- \frac{3}{2}}^{2} (2 x + 3) d x \Rightarrow I = - {[x^{2} + 3 x]}_{- 2}^{\frac{- 3}{2}} + {[x^{2} + 3 x]}_{- \frac{3}{2}}^{2} \Rightarrow I = - \frac{9}{4} + \frac{9}{2} + 4 - 6 + 4 + 6 - \frac{9}{4} + \frac{9}{2} \Rightarrow I = \frac{25}{2}$

(v)
$\int_{0}^{2} |x^{2} - 3 x + 2| d x We know that, |x^{2} - 3 x + 2| = \{\begin{cases} - (x^{2} - 3 x + 2), (x - 1) (x - 2) \leq 0 or, 1 \leq x \leq 2 \\ (x^{2} - 3 x + 2), x^{2} - 3 x + 2 \geq 0 or, x \in (- \infty, 1) \cup (2, \infty) \end{cases} ∴ I = \int_{0}^{2} (x^{2} - 3 x + 2) d x \Rightarrow I = \int_{0}^{1} (x^{2} - 3 x + 2) d x - \int_{1}^{2} (x^{2} - 3 x + 2) d x \Rightarrow I = {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{0}^{1} - {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{1}^{2} \Rightarrow I = \frac{1}{3} - \frac{3}{2} + 2 - [\frac{8}{3} - 6 + 4 - \frac{1}{3} + \frac{3}{2} - 2] \Rightarrow I = \frac{1}{3} - \frac{3}{2} + 2 - \frac{8}{3} + 6 - 2 + \frac{1}{3} - \frac{3}{2} \Rightarrow I = 1$

(vi)
$\int_{0}^{3} |3 x - 1| d x We know that, |3 x - 1| = \{\begin{cases} - (3 x - 1), 0 \leq x \leq \frac{1}{3} \\ (3 x - 1), \frac{1}{3} < x \leq 3 \end{cases} ∴ I = = \int_{0}^{\frac{1}{3}} - (3 x + 1) d x + \int_{\frac{1}{3}}^{0} (3 x + 1) d x \Rightarrow I = {[\frac{- 3 x^{2}}{2} - x]}_{0}^{\frac{1}{3}} + {[\frac{3 x^{2}}{2} + x]}_{\frac{1}{3}}^{3} \Rightarrow I = - \frac{1}{6} + \frac{1}{3} - 0 + \frac{27}{2} + 3 - \frac{1}{6} - \frac{1}{3} \Rightarrow I = \frac{65}{6}$

(vii)
$\int_{- 6}^{6} |x + 2| d x We know that, |x + 2| = \{\begin{cases} - (x + 2), - 6 \leq x \leq - 2 \\ x + 2, - 2 < x \leq 6 \end{cases} ∴ I = \int_{- 6}^{6} |x + 2| d x \Rightarrow I = \int_{- 6}^{- 2} - (x + 2) d x + \int_{- 2}^{6} (x + 2) d x \Rightarrow I = {[\frac{- x^{2}}{2} - 2 x]}_{- 6}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{6} \Rightarrow I = - 2 + 4 + 18 - 12 + 18 + 12 - 2 + 4 \Rightarrow I = 40$

(viii)
$\int_{- 2}^{2} |x + 1| d x We know that, |x + 1| = \{\begin{cases} - (x + 1), - 2 \leq x \leq - 1 \\ x + 1, - 1 < x \leq 2 \end{cases} ∴ I = \int_{- 2}^{2} |x + 1| d x \Rightarrow I = \int_{- 2}^{- 1} - (x + 1) d x + \int_{- 1}^{2} (x + 1) d x \Rightarrow I = {[\frac{- x^{2}}{2} - x]}_{- 2}^{- 1} + {[\frac{x^{2}}{2} + x]}_{- 1}^{2} \Rightarrow I = \frac{- 1}{2} + 1 + 2 - 2 + 2 + 2 - \frac{1}{2} + 1 \Rightarrow I = 5$

(ix)
$\int_{1}^{2} |x - 3| d x We know that, |x + 1| = \{\begin{cases} - (x + 1), 1 \leq x \leq 3 \\ (x + 1), x > 3 \end{cases} ∴ I = \int_{1}^{2} |x - 3| d x \Rightarrow I = \int_{1}^{2} - (x - 3) d x \Rightarrow I = {[\frac{- x^{2}}{2} - 3 x]}_{1}^{2} \Rightarrow I = - 2 - 6 + \frac{1}{2} + 3 \Rightarrow I = \frac{3}{2}$

(x)
$\int_{0}^{\frac{π}{2}} |\cos 2 x| d x We know that, |\cos 2 x| = \{\begin{cases} - \cos 2 x, \frac{π}{4} \leq x \leq \frac{π}{2} \\ \cos 2 x, 0 < x \leq \frac{π}{4} \end{cases} ∴ I = \int_{- 2}^{2} |\cos 2 x| d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \cos 2 x d x - \int_{\frac{π}{4}}^{\frac{π}{2}} \cos 2 x d x \Rightarrow I = {[\frac{\sin 2 x}{2}]}_{0}^{\frac{π}{4}} - {[\frac{\sin 2 x}{2}]}_{\frac{π}{4}}^{\frac{π}{2}} \Rightarrow I = \frac{1}{2} - 0 - 0 + \frac{1}{2} \Rightarrow I = 1$

(xi)
$\int_{0}^{2 π} |\sin x| d x We know that, |\sin x| = \{\begin{cases} - \sin x, π \leq x \leq 2 π \\ \sin x, 0 < x \leq π \end{cases} ∴ I = \int_{0}^{2 π} |\sin x| d x \Rightarrow I = \int_{0}^{π} \sin x d x + \int_{π}^{2 π} - \sin x d x \Rightarrow I = - {[\cos x]}_{0}^{π} + {[\cos x]}_{π}^{2 π} \Rightarrow I = 1 + 1 + 1 - (- 1) \Rightarrow I = 4$

(xii)
$\int_{- \frac{π}{4}}^{\frac{π}{4}} |\sin x| d x We know that, |\sin x| = \{\begin{cases} - \sin x, - \frac{π}{4} \leq x \leq 0 \\ \sin x, 0 < x \leq \frac{π}{4} \end{cases} ∴ I = \int_{- \frac{π}{4}}^{\frac{π}{4}} |\sin x| d x \Rightarrow I = \int_{- \frac{π}{4}}^{0} - \sin x d x + \int_{0}^{\frac{π}{4}} \sin x d x \Rightarrow I = {[\cos x]}_{\frac{- π}{4}}^{0} - {[\cos x]}_{0}^{\frac{- π}{4}} \Rightarrow I = 1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1 \Rightarrow I = 2 - \frac{2}{\sqrt{2}} \Rightarrow I = 2 - \sqrt{2}$

(xiii)
$\int_{2}^{8} |x - 5| d x We know that, |x - 5| = \{\begin{cases} - (x - 5), 2 \leq x \leq 5 \\ x - 5, 5 < x \leq 8 \end{cases} ∴ I = \int_{2}^{8} |x - 5| d x \Rightarrow I = \int_{2}^{5} - (x - 5) d x + \int_{5}^{8} (x - 5) d x \Rightarrow I = - {[\frac{x^{2}}{2} - 5 x]}_{2}^{5} + {[\frac{x^{2}}{2} - 5 x]}_{5}^{8} \Rightarrow I = \frac{- 25}{2} + 25 + 2 - 10 + 32 - 40 - \frac{25}{2} + 25 \Rightarrow I = 9$

(xiv)
$\int_{\frac{- π}{2}}^{\frac{π}{2}} \{\sin |x| + \cos |x|\} d x Since, f (- x) = \sin |- x| + \cos |- x| = \sin |x| + \cos |x| = f (x) S o, f (x) is an even function . ∴ I = 2 \int_{0}^{\frac{π}{2}} (\sin x + \cos x) dx \Rightarrow I = 2 {[- \cos x + \sin x]}_{0}^{\frac{π}{2}} \Rightarrow I = 2 (0 + 1 + 1 - 0) \Rightarrow I = 4$

(xv)
$\int_{0}^{4} |x - 1| d x We know that, |x - 1| = \{\begin{cases} - (x - 1), 0 \leq x \leq 1 \\ x - 1, 1 < x \leq 4 \end{cases} ∴ I = \int_{0}^{4} |x - 1| d x \Rightarrow I = \int_{0}^{1} - (x - 1) d x + \int_{1}^{4} (x - 1) d x \Rightarrow I = {[- \frac{x^{2}}{2} + x]}_{0}^{1} + {[\frac{x^{2}}{2} - x]}_{1}^{4} \Rightarrow I = \frac{- 1}{2} + 1 - 0 + 8 - 4 - \frac{1}{2} + 1 \Rightarrow I = 5$

(xvi)
$I = \int_{1}^{4} \{|x - 1| + |x - 2| + |x - 4|\} d x \Rightarrow I = \int_{1}^{4} |x - 1| d x + \int_{1}^{4} |x - 2| d x + \int_{1}^{4} |x - 4| d x We know that, |x - 1| = \{\begin{cases} - (x - 1), x \leq 1 \\ x - 1, 1 < x \leq 4 \end{cases} |x - 2| = \{\begin{cases} - (x - 2), 1 \leq x \leq 2 \\ x - 2, 2 < x \leq 4 \end{cases} |x - 4| = \{\begin{cases} - (x - 4), 1 \leq x \leq 4 \\ x - 4, x > 4 \end{cases} ∴ I = \int_{1}^{4} (x - 1) d x - \int_{1}^{2} (x - 2) d x + \int_{2}^{4} (x - 2) d x - \int_{1}^{4} (x - 4) d x \Rightarrow I = {[\frac{x^{2}}{2} - x]}_{1}^{4} - {[\frac{x^{2}}{2} - 2 x]}_{1}^{2} + {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} - {[\frac{x^{2}}{2} - 4 x]}_{1}^{4} \Rightarrow I = 8 - 4 - \frac{1}{2} + 1 - (2 - 4 - \frac{1}{2} + 2) + 8 - 8 - 2 + 4 - (8 - 16 - \frac{1}{2} + 4) \Rightarrow I = \frac{23}{2}$

(xvii)

$I = \int_{- 5}^{0} \{|x| + |x + 2| + |x + 5|\} d x \Rightarrow I = \int_{- 5}^{0} |x| d x + \int_{- 5}^{0} |x + 2| d x + \int_{- 5}^{0} |x + 5| d x We know that, |x| = \{\begin{cases} - x, - 5 \leq x \leq 0 \\ x, x > 0 \end{cases} |x + 2| = \{\begin{cases} - (x + 2), - 5 \leq x \leq - 2 \\ x + 2, - 2 < x \leq 0 \end{cases} |x + 5| = \{\begin{cases} - (x + 5), - 5 \leq x \leq 0 \\ x + 5, x > - 5 \end{cases} ∴ I = - \int_{- 5}^{0} x d x - \int_{- 5}^{- 2} (x + 2) d x + \int_{- 2}^{0} (x + 2) d x + \int_{- 5}^{0} (x + 5) d x \Rightarrow I = - {[\frac{x^{2}}{2}]}_{- 5}^{0} - {[\frac{x^{2}}{2} + 2 x]}_{- 5}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{0} + {[\frac{x^{2}}{2} + 5 x]}_{- 5}^{0} \Rightarrow I = \frac{25}{2} - (2 - 4 - \frac{25}{2} + 10) - 2 + 4 + (- \frac{25}{2} + 25) \Rightarrow I = \frac{63}{2}$

(xviii)
$I = \int_{0}^{4} \{|x| + |x - 2| + |x - 4|\} d x \Rightarrow I = \int_{0}^{4} |x| d x + \int_{0}^{4} |x - 2| d x + \int_{0}^{4} |x - 4| d x We know that, |x| = \{\begin{cases} - x, - 5 \leq x \leq 0 \\ x, x > 0 \end{cases} |x - 2| = \{\begin{cases} - (x - 2), 0 \leq x \leq 2 \\ x - 2, 2 < x \leq 4 \end{cases} |x - 4| = \{\begin{cases} - (x - 4), 0 \leq x \leq 4 \\ x - 4, x > 4 \end{cases} ∴ I = \int_{0}^{4} x d x - \int_{0}^{2} (x - 2) d x + \int_{2}^{4} (x - 2) d x - \int_{0}^{4} (x - 4) d x \Rightarrow I = {[\frac{x^{2}}{2}]}_{0}^{4} - {[\frac{x^{2}}{2} - 2 x]}_{0}^{2} + {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} - {[\frac{x^{2}}{2} - 4 x]}_{0}^{4} \Rightarrow I = 8 - (2 - 4) + 8 - 8 - 2 + 4 - (8 - 16) \Rightarrow I = 20$

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