Question

How do you find the area of the region under the curve $y=\dfrac{1}{\sqrt{2x-1}}$ from $x=\dfrac{1}{2}$ to $x=1$ ?

Answer 1

To solve these questions where you have to calculate the area of the region enclosed by a curve and two end points apply the area formula $A=\int\limits_{a}^{b}{f\left( x \right)dx}$ , where $A$ is the area under the curve, $a$ and $b$ are the lower and upper limit, respectively. After substituting the values correctly, just simply integrate the function under the integral sign and apply the limits to get the answer.

Complete step by step solution:
The given curve is: $y=\dfrac{1}{\sqrt{2x-1}}$ and the two $x$ - coordinates are $x=\dfrac{1}{2}$ and $x=1$ .
The area of the region under the curve can be calculated and given by the following formula, that is,
$A=\int\limits_{a}^{b}{f\left( x \right)dx}$ , where $A$ is the area under the curve, $a$ and $b$ are the lower and upper limit, respectively.
So, substituting the given values from the question in the above formula, we get,
$\Rightarrow A=\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{1}{\sqrt{2x-1}}dx}$ …$\left( i \right)$
Now, we can simplify the above integral by using the substitution method. Therefore, let
$u=2x-1$
Differentiating the above, we get,
$\Rightarrow \dfrac{du}{dx}=2$
Since, $1$ is a constant, therefore its differentiation is zero.
Now, since we have used the substitution method to simplify the integral, we must also change the limits of integration.
Therefore,
When $x=\dfrac{1}{2}$
$u=2\times \dfrac{1}{2}-1$
$\Rightarrow u=2\times \dfrac{1}{2}-1=0$ , and, when $x=1$ then the value of $u$ is, $u=2\times 1-1$
$\Rightarrow u=2-1=1$ .
Therefore, when $x=\dfrac{1}{2}\Rightarrow u=0$ and when $x=1\Rightarrow u=1$ .
Now, substituting all the values in the equation $\left( i \right)$ we get,
$\Rightarrow A=\dfrac{1}{2}\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{1}{\sqrt{2x-1}}\left( 2 \right)dx}$
Now, changing the limits,
$\Rightarrow A=\dfrac{1}{2}\int\limits_{0}^{1}{\dfrac{1}{\sqrt{u}}}du$
Apply the rule for the integration of a variable with power, to get,
$\Rightarrow A=\dfrac{1}{2}\left[ 2\sqrt{u} \right]_{0}^{1}$
Cancelling the like terms, that is, $2$ from the numerator and denominator after taking out it as constant and applying the higher and the lower limit, we get,
$\Rightarrow A=\sqrt{1}-\sqrt{0}$
$\Rightarrow A=1$
Simplifying the above expression, we get the area of the region under the curve $y=\dfrac{1}{\sqrt{2x-1}}$ as $A=1units$.

Note: While using the substitution method to solve integrals, always keep in mind to change the limits immediately. If after calculations the area of the region comes out to be negative then just drop the negative sign, since we know that area can never be negative. Just take its modulus and the answer will be correct.