Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How do you evaluate the definite integral $\int{\left( 2x-1 \right)dx}$ from [1,3]?

Last updated date: 22nd Jul 2024
Total views: 384.6k
Views today: 3.84k
Verified
384.6k+ views
Hint: This is a problem from definite integral. Whenever, it is asked to find the definite integral of any integration, and then we have to put the given limits after the integration in the integrated value. So, in this question, first we will find the integration and after that we will put the limits. We are going to use a formula in the solution of this question which is $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ .

Let us solve the question.
In this question, we have asked to find the definite integral (2x-1)dx from [1,3] or we can say we have to find$\int\limits_{1}^{3}{\left( 2x-1 \right)dx}$.
So, first let us find out the integration of (2x-1)dx
Let,$I=\int\limits_{1}^{3}{\left( 2x-1 \right)dx}$.
$\Rightarrow I=\int\limits_{1}^{3}{2xdx}-\int_{1}^{3}{1dx}$
Which is also can be written as
$\Rightarrow I=2\int\limits_{1}^{3}{xdx}-\int_{1}^{3}{1dx}$
As we know that any number or variable with power zero is always 1.
So, the above integration can be written as
$\Rightarrow I=2\int\limits_{1}^{3}{{{x}^{1}}dx}+\int_{1}^{3}{{{x}^{0}}dx}$
Now, using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$, we will integrate both the terms. According to this formula, we can say that the value of n is 1 in the first term and the value of n is 0 in the second term. So, the above integration can written as
$\Rightarrow I=2\left[ \dfrac{{{x}^{1+1}}}{1+1} \right]_{1}^{3}-\left[ \dfrac{{{x}^{0+1}}}{0+1} \right]_{1}^{3}$
We have put the limits here in which the lower limit is 1 and the upper limit is 3. Because it is a definite integral, so we have put the limits.
Now, we can write the above equation as
$\Rightarrow I=2\left[ \dfrac{{{x}^{2}}}{2} \right]_{1}^{3}-\left[ \dfrac{{{x}^{1}}}{1} \right]_{1}^{3}$
$\Rightarrow I=2\left[ \dfrac{{{x}^{2}}}{2} \right]_{1}^{3}-\left[ {{x}^{1}} \right]_{1}^{3}$
After putting the limiting values, we get
$\Rightarrow I=2\left[ \dfrac{{{3}^{2}}}{2}-\dfrac{{{1}^{2}}}{2} \right]-\left[ {{3}^{1}}-{{1}^{1}} \right]$
Now, let us solve the above.
$\Rightarrow I=2\left[ \dfrac{{{3}^{2}}-{{1}^{2}}}{2} \right]-\left[ {{3}^{1}}-{{1}^{1}} \right]=2\left[ \dfrac{9-1}{2} \right]-\left[ 3-1 \right]=2\times \dfrac{8}{2}-2$
$\Rightarrow I=8-2=6$
Hence, the definite integral of $\int{\left( 2x-1 \right)dx}$ from [1,3] is 6.

Note:
We should have a better knowledge in definite integrals for solving this type of question. And always remember the formulas of integration. One formula, we have used here which is very necessary to be remembered is $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$. And also don’t forget to use the limits if it is asked for the definite integral. Otherwise, the solution will be wrong.