Question

How do you find the antiderivative of \[{x^{\dfrac{3}{2}}}\]?

Answer 1

Hint: We use the definition of anti-derivatives and use that definition to find the integration of the given values. Use the general formula of integration to integrate the given value.
* Anti-derivative of a function is the opposite of the derivative of a function i.e. it is that value which can be obtained when taking reverse of the derivative function. It is that function whose derivative we take and then we get the function.
* If \[y = {x^n}\] then integration of y with respect to x will be given as \[\int {ydx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]

Complete step-by-step solution:
We are given the function \[{x^{\dfrac{3}{2}}}\]
Let us assume that \[y = {x^{\dfrac{3}{2}}}\].................… (1)
Then to calculate the antiderivative of the given function we have to calculate the integration (i.e. anti-derivative or opposite of derivative) of the function.
Since we know the opposite of derivative is integration, we will use integration.
Apply integration on both sides of equation (1)
\[ \Rightarrow \int y dx = \int {{x^{\dfrac{3}{2}}}} dx\]
Now we use the general formula of integration on right hand side of the equation
\[ \Rightarrow \int y dx = \dfrac{{{x^{\dfrac{3}{2} + 1}}}}{{\dfrac{3}{2} + 1}} + C\]
Solve the value in the power in right hand side of the equation
\[ \Rightarrow \int y dx = \dfrac{{{x^{\dfrac{5}{2}}}}}{{\dfrac{5}{2}}} + C\]
Bring the constant value in the denominator out with multiplication to the obtained value in terms of x.
\[ \Rightarrow \int y dx = \dfrac{2}{5}{x^{\dfrac{5}{2}}} + C\]
Here C is a constant of integration.

\[\therefore \]The anti-derivative of the function \[{x^{\dfrac{3}{2}}}\] is \[\dfrac{2}{5}{x^{\dfrac{5}{2}}} + C\]

Note: Many students make the mistake of calculating the derivative first and then integrate the obtained derivative which is wrong, keep in mind we have to calculate the integration of the function, not the derivative. Anti-derivative of a function means that value which on differentiation or derivative gives the function.