Question

How do you integrate \[\int {\left( {1 + 3t} \right){t^2}} dt\]?

Answer 1

Hint: To integrate \[\int {\left( {1 + 3t} \right){t^2}} dt\], we will first simplify this by multiplying the term outside the bracket with the terms inside the bracket. Then as we can see this is an indefinite integral, so we will use the formula of integration for this type i.e., \[\int {{t^n}} dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + C\], where \[n \ne - 1\]. Using this and on simplification we get the result.

Complete step by step answer:
We have to integrate \[\int {\left( {1 + 3t} \right){t^2}} dt\].
Let \[I = \int {\left( {1 + 3t} \right){t^2}} dt\].
On multiplying the terms outside the bracket with the terms inside the bracket, we get
\[ \Rightarrow I = \int {\left( {{t^2} + 3{t^3}} \right)} dt\]
Now, multiplying \[dt\] with the terms inside the bracket, we get
\[ \Rightarrow I = \int {{t^2}dt + \int {3{t^3}dt} } \]

As we know that \[\int {{t^n}} dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + C\], where \[n \ne - 1\].
Using this, we get
\[ \Rightarrow I = \dfrac{{{t^{2 + 1}}}}{{2 + 1}} + 3 \times \left( {\dfrac{{{t^{3 + 1}}}}{{3 + 1}}} \right) + C\]
On simplifying, we get
\[ \Rightarrow I = \dfrac{{{t^3}}}{3} + 3 \times \left( {\dfrac{{{t^4}}}{4}} \right) + C\]
\[ \therefore I = \dfrac{{{t^3}}}{3} + \dfrac{{3{t^4}}}{4} + C\]

Therefore, \[\int {\left( {1 + 3t} \right){t^2}} dt\] is \[\dfrac{{{t^3}}}{3} + \dfrac{{3{t^4}}}{4} + C\].

Additional information: Antidifferentiation or indefinite integration is the process of solving for an antiderivative and its opposite operation i.e., the process of finding a derivative is called differentiation. Antiderivative is generally used to evaluate the indefinite integral of any function but it can also be used to compute definite integrals by using the fundamental theorem of calculus.

The linearity of integration basically removes the complex integrand into simpler ones and hence becomes easier in case of definite integrals where limits are provided by just substituting at the place of dependent variable to get the definite answer.

Note: Here, \[C\] is the constant of integration. In a definite integral we get the value of integration constant by using the upper and lower limit of integration. Note, we can also use inverse trigonometric properties to find the antiderivative. Using the direct integration formula that is mentioned as identity can make the problem simple by just substituting those values.