\int^{16}_{1} \dfrac{x - 7}{\sqrt{x}} dx \int^{16}_{1} (x - 7)(x^{1/2}) dx :confused: Next move?
J Jason76 Senior Member Joined Oct 19, 2012 Messages 1,180 Oct 24, 2013 #1 \(\displaystyle \int^{16}_{1} \dfrac{x - 7}{\sqrt{x}} dx\) \(\displaystyle \int^{16}_{1} (x - 7)(x^{1/2}) dx\) Next move?
\(\displaystyle \int^{16}_{1} \dfrac{x - 7}{\sqrt{x}} dx\) \(\displaystyle \int^{16}_{1} (x - 7)(x^{1/2}) dx\) Next move?
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Oct 24, 2013 #2 Jason76 said: \(\displaystyle \int^{16}_{1} \dfrac{x - 7}{\sqrt{x}} dx\) \(\displaystyle \int^{16}_{1} (x - 7)(x^{1/2}) dx\) Next move? Click to expand... Multiply out. Integrate term by term.
Jason76 said: \(\displaystyle \int^{16}_{1} \dfrac{x - 7}{\sqrt{x}} dx\) \(\displaystyle \int^{16}_{1} (x - 7)(x^{1/2}) dx\) Next move? Click to expand... Multiply out. Integrate term by term.
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,251 Oct 24, 2013 #3 Jason76 said: \(\displaystyle \int^{16}_{1} \dfrac{x - 7}{\sqrt{x}} dx\) \(\displaystyle \int^{16}_{1} (x - 7)(x^{1/2}) dx\) Next move? Click to expand... \(\displaystyle \int^{16}_{1} (x - 7)(x^{-1/2}) dx \ \ \ \) You forgot to make the exponent negative as I showed here.
Jason76 said: \(\displaystyle \int^{16}_{1} \dfrac{x - 7}{\sqrt{x}} dx\) \(\displaystyle \int^{16}_{1} (x - 7)(x^{1/2}) dx\) Next move? Click to expand... \(\displaystyle \int^{16}_{1} (x - 7)(x^{-1/2}) dx \ \ \ \) You forgot to make the exponent negative as I showed here.