How do you evaluate the definite integral #int1/(x^2sqrtx)# from #[1,4]#?

Question

Paisley Johnson · Accepted Answer

#7/12# #int_a^b# #1/(x^2.sqrt2)# must be written as an integral of a power of x #sqrt2#=#x^(1/2)# So the denominator is #x ^(5/2)# So the function can be written as #x^(-5/2)# #int_1^4# #x^(-5/2)#.dx = [#-2/3x^(-3/2)#] from 4 to 1 When #x#=4 #x^(-3/2)#=#1/8# #-2/3#x#1/8#=#-1/12# When #x#=1 #x^(-3/2)# =1 #-2/3# x 1=#-2/3# #-1/12#-(#-2/3#)=#7/12#

Charles Arocha · Answer

undefined To evaluate the definite integral $ \int_{1}^{4} \frac{1}{x^2 \sqrt{x}} \, dx $, we can start by rewriting the integrand:

$ \frac{1}{x^2 \sqrt{x}} = x^{-\frac{5}{2}} $

Now, we integrate $ x^{-\frac{5}{2}} $ with respect to $ x $ from $ 1 $ to $ 4 $:

$ \int_{1}^{4} x^{-\frac{5}{2}} \, dx = \left[ \frac{x^{-\frac{3}{2}}}{-\frac{3}{2}} \right]_{1}^{4} $

Now plug in the limits:

$ = \left( \frac{4^{-\frac{3}{2}}}{-\frac{3}{2}} \right) - \left( \frac{1^{-\frac{3}{2}}}{-\frac{3}{2}} \right) $

Simplify:

$ = \left( \frac{1}{-2 \times 4^{\frac{3}{2}}} \right) - \left( \frac{1}{-2 \times 1^{\frac{3}{2}}} \right) $

$ = \left( \frac{1}{-2 \times 8} \right) - \left( \frac{1}{-2 \times 1} \right) $

$ = \left( \frac{1}{-16} \right) - \left( \frac{1}{-2} \right) $

$ = -\frac{1}{16} + \frac{1}{2} $

Finally, combine:

$ = \frac{1}{2} - \frac{1}{16} $

$ = \frac{8}{16} - \frac{1}{16} $

$ = \frac{7}{16} $

Therefore, the value of the definite integral $ \int_{1}^{4} \frac{1}{x^2 \sqrt{x}} \, dx $ is $ \frac{7}{16} $.