Question

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

Answer 1

Hint : Here, we are given with an integral and also an interval. We have to evaluate the value of the definite integral in the given interval. For this we will use the first fundamental theorem of calculus, which states that if $f$ is continuous on $[a,b]$ and $F$ is the indefinite integral of $f$ on $[a,b]$ , then $\int\limits_a^b {f(x)dx} = F(b) - F(a)$ .
Formulae used:
${\int x ^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$

Complete step-by-step answer :
We have to evaluate the definite integral $\int {\left( {2x + 2} \right)dx}$ from $[ - 3,1]$ , which can be written as $\int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx}$ .
Now, by using the fundamental theorem of calculus we get $\int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx} = F(1) - F( - 3)$ , where $F$ is the indefinite integral $\int {\left( {2x + 2} \right)dx}$ .
So, now we will integrate the integral first and then apply the limits.
$\int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx} = \left[ {\dfrac{{2{x^{1 + 1}}}}{{1 + 1}} + 2x} \right]_{ - 3}^1$ (Using ${\int x ^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$ ).
$\Rightarrow \int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx} = \left[ {\dfrac{{2{x^2}}}{2} + 2x} \right]_{ - 3}^1$
$\Rightarrow \int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx} = \left[ {{x^2} + 2x} \right]_{ - 3}^1$
$\Rightarrow \int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx} = \left( {{1^2} + 2(1)} \right) - \left( {{{( - 3)}^2} + 2( - 3)} \right)$ (Using the first fundamental theorem of calculus)
$\Rightarrow \int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx} = \left( 3 \right) - \left( {9 - 6} \right)$
$\Rightarrow \int\limits_{ - 3}^1 {\left( {2x + 2} \right)dx} = 3 - 9 + 6 = 0$
Hence, the value of the definite integral $\int {\left( {2x + 2} \right)dx}$ from $[ - 3,1]$ is 0.
So, the correct answer is “0”.

Note : If the definite integral of a function is zero, it means that the total algebraic sum of the area under the curve of the function is zero. But this doesn’t mean that the geometric sum of the area is zero. The areas under the function's curve above the x-axis and below the x-axis are equal.