Question Number 12303 by Gaurav3651 last updated on 18/Apr/17
$${What}\:{is}\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mid\mathrm{1}−{x}^{\mathrm{4}} \:\mid{dx}\:{equal}\:{to}? \\ $$$$\left({a}\right)\:\:−\mathrm{232}/\mathrm{5} \\ $$$$\left({b}\right)\:\:−\mathrm{116}/\mathrm{5} \\ $$$$\left({c}\right)\:\:\:\mathrm{116}/\mathrm{5} \\ $$$$\left({d}\right)\:\:\:\mathrm{232}/\mathrm{5} \\ $$
Answered by ajfour last updated on 18/Apr/17
![= ∫_( 1) ^3 (x^4 −1)dx =[(x^5 /5)−x]_( 1) ^( 3) =((3^5 /5)−3)−((1/5)−1) =((243)/5)−(1/5)−3+1 = ((242)/5) − 2 = ((232)/5) . (d) .](Q12310.png)
$$\:=\:\underset{\:\:\:\mathrm{1}} {\overset{\mathrm{3}} {\int}}\left({x}^{\mathrm{4}} −\mathrm{1}\right){dx}\:=\left[\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−{x}\underset{\:\:\mathrm{1}} {\overset{\:\mathrm{3}} {\right]}} \\ $$$$\:\:\:=\left(\frac{\mathrm{3}^{\mathrm{5}} }{\mathrm{5}}−\mathrm{3}\right)−\left(\frac{\mathrm{1}}{\mathrm{5}}−\mathrm{1}\right) \\ $$$$\:\:\:=\frac{\mathrm{243}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{5}}−\mathrm{3}+\mathrm{1} \\ $$$$\:\:\:=\:\frac{\mathrm{242}}{\mathrm{5}}\:−\:\mathrm{2}\:=\:\frac{\mathrm{232}}{\mathrm{5}}\:.\:\:\left(\boldsymbol{{d}}\right)\:. \\ $$
Commented by prakash jain last updated on 22/Apr/17
![For module problem you need to approach in the following way ∣f(x)∣ find range of x for which f(x)≥0 ∣f(x)∣ find range of x for which f(x)<0 Split the definite integral in parts where f(x)≥0 and where f(x)<0 replace ∣f(x)∣ by f(x) or −f(x) For example ∫_0 ^3 ∣1−x^2 ∣ dx ∣1−x^2 ∣≥0 for x∈[0,1] ∣1−x^2 ∣<0 for x∈(1,3] ∫_0 ^3 ∣1−x^2 ∣ dx=∫_0 ^1 (1−x^2 )dx+∫_1 ^2 −(1−x^2 )dx](Q12427.png)
$$\mathrm{For}\:\mathrm{module}\:\mathrm{problem}\:\mathrm{you}\:\mathrm{need} \\ $$$$\mathrm{to}\:\mathrm{approach}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{way} \\ $$$$\mid{f}\left({x}\right)\mid\:\mathrm{find}\:\mathrm{range}\:\mathrm{of}\:{x}\:\mathrm{for}\:\mathrm{which}\:{f}\left({x}\right)\geqslant\mathrm{0} \\ $$$$\mid{f}\left({x}\right)\mid\:\mathrm{find}\:\mathrm{range}\:\mathrm{of}\:{x}\:\mathrm{for}\:\mathrm{which}\:{f}\left({x}\right)<\mathrm{0} \\ $$$$\mathrm{Split}\:\mathrm{the}\:\mathrm{definite}\:\mathrm{integral}\:\mathrm{in}\:\mathrm{parts} \\ $$$$\mathrm{where}\:{f}\left({x}\right)\geqslant\mathrm{0}\:\mathrm{and}\:\mathrm{where}\:{f}\left({x}\right)<\mathrm{0} \\ $$$$\mathrm{replace}\:\mid{f}\left({x}\right)\mid\:\mathrm{by}\:{f}\left({x}\right)\:\mathrm{or}\:−{f}\left({x}\right) \\ $$$$\mathrm{For}\:\mathrm{example} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} \mid\mathrm{1}−{x}^{\mathrm{2}} \mid\:\mathrm{d}{x} \\ $$$$\mid\mathrm{1}−{x}^{\mathrm{2}} \mid\geqslant\mathrm{0}\:\mathrm{for}\:{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mid\mathrm{1}−{x}^{\mathrm{2}} \mid<\mathrm{0}\:\mathrm{for}\:{x}\in\left(\mathrm{1},\mathrm{3}\right] \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} \mid\mathrm{1}−{x}^{\mathrm{2}} \mid\:\mathrm{d}{x}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}+\int_{\mathrm{1}} ^{\mathrm{2}} −\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$$$ \\ $$
Commented by Gaurav3651 last updated on 19/Apr/17
$$\: \\ $$$${shouldn}'{t}\:{we}\:{proceed}\:{in}\:{different} \\ $$$${way}\:{to}\:{solve}\:{modulus}\:{problems} \\ $$$${otherwise}\:{it}\:{is}\:{similar}\:{to} \\ $$$$\underset{\:\:\:\mathrm{1}} {\overset{\mathrm{3}} {\int}}\left({x}^{\mathrm{4}} −\mathrm{1}\right){dx}\:\: \\ $$$$ \\ $$