
How do you evaluate the definite integral \[\int {2xdx} \] from \[\left[ {0,1} \right] \] ?
Answer
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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.
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