Question

Let
$f(x) = \sqrt[3]{ {x}^{2} + 4x }$
And let $\red g(x)$be an Atiderivative of $\red f(x)$ Then if $\red g(5)=7$
Find:
$\red g(1)$

Answer it correctly✔. Please don't spam!!
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Answer 1

Correct Question :-

$f(x)=\sqrt{x^2+4x}$

$g(x)=\displaystyle\int f(x)\:dx$

$g(5)=7$

$g(1)=?$

Solution :-

Calculating g(x),

$g(x)=\displaystyle\int f(x)\:dx$

$g(x)=\displaystyle\int \sqrt{x^2+4x}\:dx$

$g(x)=\displaystyle\int \sqrt{x^2+4x+4-4}\:dx$

$g(x)=\displaystyle\int \sqrt{(x+2)^2-2^2}\:dx$

$g(x)=\dfrac{x+2}{2}\sqrt{x^2+4x}-2\ln(x+2+\sqrt{x^2+4x})+C$

$\left(\because \displaystyle\int\sqrt{t^2-a^2}\:dt=\dfrac{t}{2}\sqrt{t^2-a^2}-\dfrac{a^2}{2}\ln (t+\sqrt{t^2-a^2})+C\right)$

Calculating 'C' from g(5),

$g(5)=\dfrac{5+2}{2}\sqrt{5^2+4(5)}-2\ln(5+2+\sqrt{5^2+4(5)})+C$

$g(5)=\dfrac{7}{2}\sqrt{25+20}-2\ln(7+\sqrt{25+20})+C$

$g(5)=\dfrac{7}{2}\sqrt{45}-2\ln(7+\sqrt{45})+C$

$7=\dfrac{21}{2}\sqrt{5}-2\ln(7+3\sqrt{5})+C$

$C=7-\dfrac{21}{2}\sqrt{5}+2\ln(7+3\sqrt{5})$

Writing g(x),

$g(x)=\dfrac{x+2}{2}\sqrt{x^2+4x}-2\ln(x+2+\sqrt{x^2+4x})+7-\dfrac{21}{2}\sqrt{5}+2\ln(7+3\sqrt{5})$

Calculating g(1) from g(x),

$g(1)=\dfrac{1+2}{2}\sqrt{1^2+4(1)}-2\ln(1+2+\sqrt{1^2+4(1)})+7-\dfrac{21}{2}\sqrt{5}+2\ln(7+3\sqrt{5})$

$g(1)=\dfrac{3}{2}\sqrt{5}-2\ln(3+\sqrt{5})+7-\dfrac{21}{2}\sqrt{5}+2\ln(7+3\sqrt{5})$

$g(1)=7+\dfrac{3}{2}\sqrt{5}-\dfrac{21}{2}\sqrt{5}+2\ln(7+3\sqrt{5})-2\ln(3+\sqrt{5})$

$g(1)=7-\dfrac{18}{2}\sqrt{5}+2\left(\ln(7+3\sqrt{5})-\ln(3+\sqrt{5})\right)$

$g(1)=7-9\sqrt{5}+2\ln\left(\dfrac{7+3\sqrt{5}}{3+\sqrt{5}}\right)$

$\left(\because \ln a-\ln b=\ln\dfrac{a}{b}\right)$

$g(1)=7-9\sqrt{5}+2\ln\left(\dfrac{1(3+\sqrt{5})^2}{2(3+\sqrt{5})}\right)$

$g(1)=7-9\sqrt{5}+2\ln\left(\dfrac{3+\sqrt{5}}{2}\right)$

Answer :-

$\underline{\boxed{g(1)=7-9\sqrt{5}+2\ln\left(\dfrac{3+\sqrt{5}}{2}\right)}}$