Question

How do you evaluate the definite integral \[\int{\left( x-2 \right)dx}\] from [-1,0]?

Answer 1

Hint: This type of problem is based on the concept of definite integral. First, we have to substitute the limits to the integration. Here, the interval is [-1,0], thus the definite integral is \[\int\limits_{-1}^{0}{\left( x-2 \right)dx}\]. Then, use the subtraction rule of integration \[\int{\left( a-b \right)dx=\int{adx-\int{bdx}}}\] to the given function. And find the integral using the power rule of integration that is \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}\]. Here, n=1. Do necessary calculations and substitute the limits in the variable.

Complete step by step solution:
According to the question, we are asked to find the value of the definite integral of \[\int{\left( x-2 \right)dx}\] from [-1,0].
We have been given the function is x-2 and the limits are [-1,0].
We know that the definite integral of \[\int{\left( x-2 \right)dx}\] over the limit [-1,0] will be
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}\] ----------(1)
Now, we have to solve this integration.
We know that the subtraction rule of integration is \[\int{\left( a-b \right)dx=\int{adx-\int{bdx}}}\].
Here, a=x and b=2.
Therefore, we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\int\limits_{-1}^{0}{xdx}-\int\limits_{-1}^{0}{2dx}\]
When we have a constant multiplied to a variable in the integral, we can take the constant out of the integrals.
\[\Rightarrow \int\limits_{-1}^{0}{\left( x-2 \right)dx}=\int\limits_{-1}^{0}{xdx}-2\int\limits_{-1}^{0}{dx}\]
We know that \[\int{dx=x}\]. Therefore, we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\int\limits_{-1}^{0}{xdx}-2\left[ x \right]_{-1}^{0}\]
Using the power rule of integration \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}\], we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\left[ \dfrac{{{x}^{1+1}}}{1+1} \right]_{-1}^{0}-2\left[ x \right]_{-1}^{0}\]
On further simplification, we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\left[ \dfrac{{{x}^{2}}}{2} \right]_{-1}^{0}-2\left[ x \right]_{-1}^{0}\]
On substituting the limits to the variable x, we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\left[ \dfrac{{{0}^{2}}-{{\left( -1 \right)}^{2}}}{2} \right]-2\left[ 0-\left( -1 \right) \right]\]
On further simplification, we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\left[ \dfrac{{{0}^{2}}-{{\left( -1 \right)}^{2}}}{2} \right]-2\left[ 0+1 \right]\]
We know that the square of 0 is 0 and the square of -1 is 1.
\[\Rightarrow \int\limits_{-1}^{0}{\left( x-2 \right)dx}=\left[ \dfrac{0-1}{2} \right]-2\times 1\]
\[\Rightarrow \int\limits_{-1}^{0}{\left( x-2 \right)dx}=\dfrac{-1}{2}-2\]
On taking LCM, we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\dfrac{-1-2\times 2}{2}\]
On further simplification, we get
\[\int\limits_{-1}^{0}{\left( x-2 \right)dx}=\dfrac{-1-4}{2}\]
\[\Rightarrow \int\limits_{-1}^{0}{\left( x-2 \right)dx}=\dfrac{-5}{2}\]
Therefore, the value of the definite integral \[\int{\left( x-2 \right)dx}\] from [-1,0] is \[\dfrac{-5}{2}\].

Note:
We should substitute the intervals in the definite integrals carefully. On changing the upper and lower limits, we get incorrect answers. Avoid calculation mistakes based on sign conventions. We should know the properties of integration to solve this type of question.