Question

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

Answer 1

Hint : In this question to evaluate the definite integral of the given function we first integrate the given function by using rules of integration. After doing integration we use the formula $\int\limits_a^b {f(x)} dx = F\left( b \right) - F\left( a \right)$ to solve it further. After this there is only calculation that is addition or subtraction only. On further solving it we get the value of the definite integral of a given expression.

Complete step-by-step answer :
The definite integral of a real-valued function f(x) with respect to a real variable x on an interval [a,b] is expressed as
$\int\limits_a^b {f(x)} dx = F\left( b \right) - F\left( a \right)$
Where a is lower limit, b is upper limit and $\int\limits_a^b {f(x)} dx$ is read as the definite integral of f(x) with respect to dx from a to b.
To do the integration of the functions we should know the integration rules. The integration rules are defined for different types of functions. In the given expression, basic rules of integration will be used.
$\Rightarrow \int {\left( {{t^2} + 2} \right)} dt$
In the given expression there are two types of terms inside the bracket. One is the squared term and the other is the constant term. The integration of constant is shown below
$\int {a{\text{ }}} dt = ax + C$ where a is a constant
And the integration of square term is
$\int {{x^2}{\text{ }}} dt = \dfrac{{{x^3}}}{3} + C$ where x is any variable
Now the given function of which we have to find the integral is
$\Rightarrow {}_{ - 1}^1\int {\left( {{t^2} + 2} \right)} dt$
By using sum rule of integration we can also write the above function as
$\Rightarrow {}_{ - 1}^1\int {{t^2}} dt + \int {2dt}$
By using the integration rules which we discussed above we get
$\Rightarrow \left[ {\dfrac{{{t^3}}}{3}} \right]_{ - 1}^1 + \left[ {2t} \right]_{ - 1}^1$
Now by using the formula $\int\limits_a^b {f(x)} dx = F\left( b \right) - F\left( a \right)$ we have
$\Rightarrow \left[ {\dfrac{1}{3} - \dfrac{{\left( { - 1} \right)}}{3}} \right] + \left[ {2 - \left( { - 2} \right)} \right]$
As we know that minus into minus is plus therefore
$\Rightarrow \left[ {\dfrac{1}{3} + \dfrac{1}{3}} \right] + \left[ {2 + 2} \right]$
On adding,
$\Rightarrow \dfrac{2}{3} + 4$
On further solving we get
$\Rightarrow \dfrac{{2 + 12}}{3}$
By doing addition in the numerator we get
$\Rightarrow \dfrac{{14}}{3}$
Hence, the definite integral of the given function is $\dfrac{{14}}{3}$
So, the correct answer is “$\dfrac{{14}}{3}$”.

Note : Keep in mind that the definite integral is the difference between the values of the integral of a given function f(x) for an upper value and a lower value of the independent variable x. Remember the rules of integration to evaluate the given functions.