Question

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

Answer 1

Hint: Given a definite integral expression in which we have to calculate the derivative under the integral sign. First, we will represent the integral using the lower and upper limits. Then, we will write the integral as a sum and difference of the integrals. Then, we will apply the integral rule to find the integration of the expression. Then, substitute the limits to the resultant expression and subtract the lower limit from the upper limit.

Formula used:
The integration of the variable t with respect to dt is given by:
$\int_{a\left( x \right)}^{b\left( x \right)} {tdt} = \left[ {\dfrac{{{t^2}}}{2}} \right]_a^b$
The integration of the constant term is given as:
$\int_{a\left( x \right)}^{b\left( x \right)} {dt} = \left[ t \right]_a^b$

Complete step-by-step answer:
We are given the expression. Then, first we will apply the limits to the expression by substituting $b = 3$ and $a = - 2$.
$\int_{ - 2}^3 {\left( { - 2x + 2} \right)dx} $
Now, we will write the expression as a sum of two integrals.
$\int_{ - 2}^3 {\left( { - 2x + 2} \right)dx} = \int_{ - 2}^3 {\left( { - 2x} \right)dx} + \int_{ - 2}^3 {\left( 2 \right)dx} $
Now, we will find the integration of first expression $\int_{ - 2}^3 {\left( { - 2x} \right)dx} $by taking out the constant term out of the integral.
$ \Rightarrow \int_{ - 2}^3 {\left( { - 2x} \right)dx} = - 2\int_{ - 2}^3 {\left( x \right)dx} $
Then, apply the integration formula to the expression.
$ \Rightarrow \int_{ - 2}^3 {\left( { - 2x} \right)dx} = - 2\left[ {\dfrac{{{x^2}}}{2}} \right]_{ - 2}^3$
Now, again taking the constant term out of the expression.
$ \Rightarrow \int_{ - 2}^3 {\left( { - 2x} \right)dx} = - 2 \times \dfrac{1}{2}\left[ {{x^2}} \right]_{ - 2}^3$
Cancel out the common terms, we get:
$ \Rightarrow \int_{ - 2}^3 {\left( { - 2x} \right)dx} = - \left[ {{x^2}} \right]_{ - 2}^3$ …… (1)
Now, we will find the integration of first expression $\int_{ - 2}^3 {\left( 2 \right)dx} $by taking out the constant term out of the integral.
$ \Rightarrow \int_{ - 2}^3 {\left( 2 \right)dx} = 2\int_{ - 2}^3 {\left( 1 \right)dx} $
Then, apply the integration formula to the expression.
$ \Rightarrow 2\int_{ - 2}^3 {\left( 1 \right)dx} = 2\left[ x \right]_{ - 2}^3$ …… (2)
Now, we will add the values in the equation (1) and (2).
$ - \left[ {{x^2}} \right]_{ - 2}^3 + 2\left[ x \right]_{ - 2}^3$
Now, we will substitute the value of upper limit and lower limit and subtract them.
$ - \left[ {{3^2} - {{\left( { - 2} \right)}^2}} \right] + 2\left[ {3 - \left( { - 2} \right)} \right]$
On simplifying the expression, we get:
$ \Rightarrow - \left( {9 - 4} \right) + 2\left( {3 + 2} \right)$
$ \Rightarrow - 5 + 2\left( 5 \right)$
$ \Rightarrow - 5 + 10 = 5$

Hence, the value of definite integral of the expression is $5$

Additional information: The quotient is a number that is obtained after dividing one number by another number. The number which is divided or written on the numerator is known as a dividend. The number written at the denominator of the fraction is known as a divisor.

Note:
 In such types of questions the students mainly don't get an approach on how to solve it. In such types of questions students mainly forget to apply the upper and lower limits to the expression. In such types of questions, students may get confused to find the value of the derivative of the expression.