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Question-191921




Question Number 191921 by Rupesh123 last updated on 03/May/23
Answered by AST last updated on 03/May/23
∫(3x+5)dx=((3x^2 )/2)+5x+c  ∫_(−4) ^2 (3x+5)dx=(16+c)−(4+c)=12  ⇒(4+log_3 x)(log_3 x)=12  ⇒(4+p)p=12⇒p^2 +4p−12=0     (p=log_3 x)  ⇒p=−6 or 2  ⇒x=3^(−6) =(1/(729)) or 9
$$\int\left(\mathrm{3}{x}+\mathrm{5}\right){dx}=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{5}{x}+{c} \\ $$$$\int_{−\mathrm{4}} ^{\mathrm{2}} \left(\mathrm{3}{x}+\mathrm{5}\right){dx}=\left(\mathrm{16}+{c}\right)−\left(\mathrm{4}+{c}\right)=\mathrm{12} \\ $$$$\Rightarrow\left(\mathrm{4}+{log}_{\mathrm{3}} {x}\right)\left({log}_{\mathrm{3}} {x}\right)=\mathrm{12} \\ $$$$\Rightarrow\left(\mathrm{4}+{p}\right){p}=\mathrm{12}\Rightarrow{p}^{\mathrm{2}} +\mathrm{4}{p}−\mathrm{12}=\mathrm{0}\:\:\:\:\:\left({p}={log}_{\mathrm{3}} {x}\right) \\ $$$$\Rightarrow{p}=−\mathrm{6}\:{or}\:\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{3}^{−\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{729}}\:{or}\:\mathrm{9} \\ $$
Commented by Rupesh123 last updated on 04/May/23
Perfect ��

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