How do you evaluate the definite integral \[\int {2xdx} \] from \[\left[ {0,1} \right] \] ?
Answer
587.1k+ views
Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know the basic integration formulae to solve this question. Also, we need to know the difference between definite integral and indefinite integral calculation. Also, we need to know how to apply the limit in the integral calculation.
Complete step-by-step answer:
The given question is shown below,
\[\int {2xdx} \] From \[\left[ {0,1} \right] \] ?
It also can be written as,
\[\int\limits_0^1 {2xdx} = ? \to \left( 1 \right)\]
Let’s see the difference between the definite integral and indefinite integral calculations. If there is no limit is given in the problem which is called a definite integral. Otherwise, it is known as indefinite integration. In this problem, we have definite integral.
We know that,
\[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} \to \left( 2 \right)\]
In the equation, \[\left( 1 \right)\] we have the constant term \[2\] , so it moves outside from the integration symbol. So, the equation \[\left( 1 \right)\] can also be written as,
\[
\left( 1 \right) \to \int\limits_0^1 {2xdx} = ? \\
2\int\limits_0^1 {xdx} = ? \to \left( 3 \right) \\
\]
By using the equation \[\left( 2 \right)\] , the equation \[\left( 3 \right)\] can also be written as,
\[2\int\limits_0^1 {xdx} = 2\left[ {\dfrac{{{x^2}}}{2}} \right] _0^1\]
(Here \[1\] is the upper limit and \[0\] is a lower limit)
The above equation can also be written as,
\[2\int\limits_0^1 {xdx} = 2\left[ {\dfrac{{{x^2}}}{2}} \right] _0^1 = 2 \times \dfrac{1}{2}\left[ {{x^2}} \right] _0^1\]
We would substitute the upper limit and lower limit instead of \[x\] .
\[
2\int\limits_0^1 {xdx} = 1\left[ {{x^2}} \right] _0^1 \\
2\int\limits_0^1 {xdx} = \left[ {{{\left( 1 \right)}^2} - {{\left( 0 \right)}^2}} \right] \\
\]
(We put \[ - \] symbol between the upper limit and lower limit after substituting \[x\] value)
\[2\int\limits_0^1 {xdx} = 1 - 0\]
\[2\int\limits_0^1 {xdx} = 1\]
So, the final answer is,
\[2\int\limits_0^1 {xdx} = 1\]
So, the correct answer is “1”.
Note: We would remember the basic integration formulae to solve these types of questions. If the limit is not given we would add \[C\] (constant term) in the final answer. Note that we won’t need to integrate the constant term and we can take it constantly outside while applying the limit to make the easy calculation.
Complete step-by-step answer:
The given question is shown below,
\[\int {2xdx} \] From \[\left[ {0,1} \right] \] ?
It also can be written as,
\[\int\limits_0^1 {2xdx} = ? \to \left( 1 \right)\]
Let’s see the difference between the definite integral and indefinite integral calculations. If there is no limit is given in the problem which is called a definite integral. Otherwise, it is known as indefinite integration. In this problem, we have definite integral.
We know that,
\[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} \to \left( 2 \right)\]
In the equation, \[\left( 1 \right)\] we have the constant term \[2\] , so it moves outside from the integration symbol. So, the equation \[\left( 1 \right)\] can also be written as,
\[
\left( 1 \right) \to \int\limits_0^1 {2xdx} = ? \\
2\int\limits_0^1 {xdx} = ? \to \left( 3 \right) \\
\]
By using the equation \[\left( 2 \right)\] , the equation \[\left( 3 \right)\] can also be written as,
\[2\int\limits_0^1 {xdx} = 2\left[ {\dfrac{{{x^2}}}{2}} \right] _0^1\]
(Here \[1\] is the upper limit and \[0\] is a lower limit)
The above equation can also be written as,
\[2\int\limits_0^1 {xdx} = 2\left[ {\dfrac{{{x^2}}}{2}} \right] _0^1 = 2 \times \dfrac{1}{2}\left[ {{x^2}} \right] _0^1\]
We would substitute the upper limit and lower limit instead of \[x\] .
\[
2\int\limits_0^1 {xdx} = 1\left[ {{x^2}} \right] _0^1 \\
2\int\limits_0^1 {xdx} = \left[ {{{\left( 1 \right)}^2} - {{\left( 0 \right)}^2}} \right] \\
\]
(We put \[ - \] symbol between the upper limit and lower limit after substituting \[x\] value)
\[2\int\limits_0^1 {xdx} = 1 - 0\]
\[2\int\limits_0^1 {xdx} = 1\]
So, the final answer is,
\[2\int\limits_0^1 {xdx} = 1\]
So, the correct answer is “1”.
Note: We would remember the basic integration formulae to solve these types of questions. If the limit is not given we would add \[C\] (constant term) in the final answer. Note that we won’t need to integrate the constant term and we can take it constantly outside while applying the limit to make the easy calculation.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

