g(x)=#int_sqrtx^xtan^(-1)tdt# Find?
1) g(#sqrt3#)
2) g'(#sqrt3#)
3) g"(#sqrt3#)

Question

g(x)=#int_sqrtx^xtan^(-1)tdt#   Find?
1) g(#sqrt3#)
2) g'(#sqrt3#)
3) g"(#sqrt3#)

Isaiah Allen · Accepted Answer

Please see below.

Here, #g(x)=int_sqrtx^x tan^-1tdt=int_sqrtx^x(1) tan^-1tdt# #"Using "color(blue) "Integration by Parts:"# #I=tan^-1t int1dt-int(d/(dt)(tan^-1t)int1dt)dt# #I=[tan^-1txxt]_sqrtx^x -int_sqrtx^xt/(1+t^2)dt# #=[t*tan^-1t]_sqrtx^x-1/2int_sqrtx^x(2t)/(1+t^2)dt# #=xtan^-1x-sqrtxtan^-1sqrtx-1/2[ln|1+t^2|]_sqrtx^x# #g(x)=xtan^-1x-sqrtxtan^-1sqrtx-1/2[ln|1+x^2|-ln|1+x|]# #g(x)=xtan^-1x-sqrtxtan^-1sqrtx-1/2ln|(1+x^2)/(1+x)|...to(A)# #:.g(sqrt3)=sqrt3xxpi/3-root(4)3tan^-1root(4)3-1/2ln| (1+sqrt3)/(1+root(4)3)|# From #(A)#, #g'(x)=tan^-1x-1/(2sqrtx)tan^-1sqrtx# #g'(sqrt3)=pi/3-1/(2root(4)3)tan^-1root(4)3# g"#(x)=1/(1+x^2)-1/(4x(1+x))+1/(4x^(3/2))tan^-1sqrtx# g"#(sqrt3)=1/4-1/(4sqrt3(1+sqrt3))+1/(4(3)^(3/4))tan^-1root(4)3#

Axel Alvarez · Answer

undefined To find the value of $ g(\sqrt{3}) $, we need to evaluate the definite integral $ \int_{\sqrt{3}}^{\sqrt{3}} t \tan^{-1}(t) \, dt $. Since the upper and lower limits of integration are the same, the integral evaluates to zero. Therefore, $ g(\sqrt{3}) = 0 $.

To find the first derivative $ g'(\sqrt{3}) $, we can use the Fundamental Theorem of Calculus. Differentiating under the integral sign with respect to $ x $, we get $ g'(x) = \frac{d}{dx} \int_{\sqrt{x}}^{x} t \tan^{-1}(t) \, dt $. Applying the Fundamental Theorem of Calculus, we get $ g'(x) = x \tan^{-1}(x) - \frac{1}{2} \ln(x^2 + 1) + \frac{\sqrt{x}}{2} $. Evaluating this at $ x = \sqrt{3} $, we find $ g'(\sqrt{3}) $.

To find the second derivative $ g''(\sqrt{3}) $, we differentiate $ g'(x) $ with respect to $ x $. After differentiation, we evaluate $ g''(\sqrt{3}) $ to find the second derivative at $ x = \sqrt{3} $.

g(x)=#int_sqrtx^xtan^(-1)tdt# Find? 1) g(#sqrt3#) 2) g'(#sqrt3#) 3) g"(#sqrt3#)