Question

How do you evaluate the integral $\int{\dfrac{1}{\sqrt{1-x}}dx}$ from $0$ to $1?$

Answer 1

Here, in the above given question we have to evaluate it by using integration by substitution which is also known as substitution. After integrating you have two options to evaluate for the given limits of integration. Or you can change the limits of integration.

Complete step-by-step answer:
As the given equation or integral we have,
$\Rightarrow$ $\int{\dfrac{1}{\sqrt{1-x}}dx}$
We can evaluate the integral by using integration by substitution.
So, we pick some part of the integer and set it equal to some variable (such as $'u')$ but you can also take another variable.
Now, we can write,
$\Rightarrow$ $u=1-x$
Therefore,
$\Rightarrow$ $du=-1dx$
$\Rightarrow$ $-du=dx$
Now, substitute the values in our integral we get,
$\Rightarrow$ $-\int{\dfrac{1}{\sqrt{u}}du,}$ it is also written as
$\Rightarrow$ $-\int{{{u}^{-\dfrac{1}{2}}}du}$
By integrating we get,
$\Rightarrow$ $-\int{{{u}^{-\dfrac{1}{2}}}du=-2{{u}^{\dfrac{1}{2}}}}$
Here, we have two options on evaluating for the given limits of integration.
Then, substitute $1-x$ as $'u'$ we get,
$\Rightarrow$ $-2{{\left( 1-x \right)}^{\dfrac{1}{2}}}$
$\Rightarrow$ $-2\left[ {{\left( 1-1 \right)}^{\dfrac{1}{2}}}-{{\left( 1-0 \right)}^{\dfrac{1}{2}}} \right]$
$\Rightarrow$ $-2\left( -1 \right)$
$\Rightarrow 2$
You can also find the answer another way.
By changing limit of integration:
$\Rightarrow$ $u=1-x$
$\Rightarrow$ $u=1-\left( -1 \right)$
$\Rightarrow$ $u=0$ [new upper limit]
$\Rightarrow$ $u=1-0$
$\Rightarrow$ $u=1$ [new lower limit]
$\Rightarrow$ $-2\left[ {{\left( 0 \right)}^{\dfrac{1}{2}}}-{{\left( 1 \right)}^{\dfrac{1}{2}}} \right]$
$\Rightarrow$ $-2\left( -1 \right)$
$\Rightarrow 2$

By evaluate the integral $\int{\dfrac{1}{\sqrt{1-x}}dx}$ from $0$ to $1$ is $2.$

Additional Information:
As we have a definite integral for example. We take $\int\limits_{a}^{b}{f(x)dx}$ it looks like this,
So, the definite integrals differ from the integrals that are indefinite because of the $'a'$ as a lower limit and $'b'$ as upper limits.
Then, we have to know about integration by substitution. It is used to find the integrals to some slightly trick functions. Then standard integrals. It is useful for working with functions that fall into the class for some function which multiplied by its derivative. In this type of problem if we multiply a fraction by its denominator then the given denominator disappears.

Note:
First check out the problem and also check the formula which is applied there. Sometimes we get wrong answers due to incorrect formulas. After evaluating you have to cross check the answer. Whether it is correct or not. Many times we misplaced the value along with the sign. There are key points you need to check before solving any problem.