\(\displaystyle \int_{1}^{2} \dfrac{dx}{\sqrt{x -1}}\)
Discontinuity at \(\displaystyle x = 1\)
\(\displaystyle \int_{b}^{2} \dfrac{dx}{\sqrt{x -1}} dx\)
\(\displaystyle \int_{b}^{2} (x - 1)^{-1/2} dx\)
\(\displaystyle \rightarrow \dfrac{(x - 1)^{1/2}}{\dfrac{1}{2}}\) - evaluated at \(\displaystyle b\) (lower bound) and \(\displaystyle 2\) (upper bound)
\(\displaystyle \rightarrow (2)(x - 1)^{1/2}\) - evaluated at \(\displaystyle b\) (lower bound) and \(\displaystyle 2\) (upper bound)
Now upperbound \(\displaystyle -\) lower bound to find the definite integral.
\(\displaystyle [(2)((2) - 1)^{1/2}] - [(2)((b) - 1)^{1/2}] \)
\(\displaystyle [(2)(1)^{1/2}] - [(2)((b) - 1)^{1/2}] \)
\(\displaystyle [2] - [(2)((b) - 1)^{1/2}] \)
\(\displaystyle \lim b \rightarrow 1^{+}[[2] - [(2)((b) - 1)^{1/2}]] \)
\(\displaystyle \lim b \rightarrow 1^{+}[[2] - [(2)((1) - 1)^{1/2}]] = 2 \) - Because it has a limit, it converges. It's area can be taken, unlike if it diverged.
Is this right. Any error?
Discontinuity at \(\displaystyle x = 1\)
\(\displaystyle \int_{b}^{2} \dfrac{dx}{\sqrt{x -1}} dx\)
\(\displaystyle \int_{b}^{2} (x - 1)^{-1/2} dx\)
\(\displaystyle \rightarrow \dfrac{(x - 1)^{1/2}}{\dfrac{1}{2}}\) - evaluated at \(\displaystyle b\) (lower bound) and \(\displaystyle 2\) (upper bound)
\(\displaystyle \rightarrow (2)(x - 1)^{1/2}\) - evaluated at \(\displaystyle b\) (lower bound) and \(\displaystyle 2\) (upper bound)
Now upperbound \(\displaystyle -\) lower bound to find the definite integral.
\(\displaystyle [(2)((2) - 1)^{1/2}] - [(2)((b) - 1)^{1/2}] \)
\(\displaystyle [(2)(1)^{1/2}] - [(2)((b) - 1)^{1/2}] \)
\(\displaystyle [2] - [(2)((b) - 1)^{1/2}] \)
\(\displaystyle \lim b \rightarrow 1^{+}[[2] - [(2)((b) - 1)^{1/2}]] \)
\(\displaystyle \lim b \rightarrow 1^{+}[[2] - [(2)((1) - 1)^{1/2}]] = 2 \) - Because it has a limit, it converges. It's area can be taken, unlike if it diverged.
Is this right. Any error?
Last edited: