Question

What is the integral of $$e^{x^3}$$?

Answer 1

Hint: In order to determine the integral of the given exponential function. In mathematics, an integral assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinitesimal data. The process of finding integrals is called integration.

Complete step by step solution:
We are given the exponential function is $$\int e^{x^3}dx$$
Now, Let us consider, $$z=x^3$$
Differentiating the exponential o ‘z’ with respect to ‘x’ , then
$$\begin{gathered} \Rightarrow 3x^{2}dx=dz \\ \Rightarrow dx={\displaystyle \frac{dz}{3x^2}} \end{gathered}$$
Comparing the exponential function $$z=x^n$$ with $$dx={\displaystyle \frac{1}{n}}{z}^{{\displaystyle \frac{1}{n}}-1}dz$$, so we can write
$$\begin{gathered} dx={\displaystyle \frac{1}{3}}{z}^{1-3}dz \\ dx={\displaystyle \frac{1}{3}}{z}^{{\displaystyle \frac{1}{3}}-1}dz \end{gathered}$$
Since, $$n=3$$
Now we can substitute the ‘dx’ and ‘z’ value into the given equation
$$\int e^{x^3}dx=\int e^z{\displaystyle \frac{dz}{3x^2}}$$$$\int e^{x^3}dx=\int e^z{\displaystyle \frac{dz}{3x^2}}$$
We take the integral limit as $$0$$to $$\mathrm{\infty }$$, we get
$$\int e^{x^3}dx=\int e^z{\displaystyle \frac{1}{3}}{z}^{{\displaystyle \frac{1}{3}}-1}dz$$
$$\int e^{x^3}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^z{\displaystyle \frac{1}{3}}{z}^{{\displaystyle \frac{1}{3}}-1}dz$$
Expand the integral values on the exponential function, we can get
$$\int e^{x^3}dx={\displaystyle \frac{1}{3}}\left(\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^z{z}^{{\displaystyle \frac{1}{3}}-1}dz-\underset{z}{\overset{\mathrm{\infty }}{\int }}{e}^z{z}^{{\displaystyle \frac{1}{3}}-1}dz\right)+c$$
On compare the formula for indefinite integral $$\mathrm{\Gamma }(n,z)+d$$ with the above derivative equation, the
$$\int e^{x^3}dx={\displaystyle \frac{1}{3}}\mathrm{\Gamma }({\displaystyle \frac{1}{3}},x^3)+d$$
Where $d$ and $c$ are constant.
Hence, the integral of $$e^{x^3}$$is $${\displaystyle \frac{1}{3}}\mathrm{\Gamma }({\displaystyle \frac{1}{3}},x^3)+d$$.

Additional information:
In integral, there are two types of integrals in maths:
> Definite Integral
> Indefinite Integral
Definite Integral:
An integral that contains the upper and lower limits then it is a definite integral. On a real line, x is restricted to lie. Riemann Integral is the other name of the Definite Integral.
A definite Integral is represented as:
$$\underset{a}{\overset{b}{\int }}f(x)dx$$ Indefinite Integral:
Indefinite integrals are defined without upper and lower limits. It is represented as:
$$\int f(x)dx=F(x)+C$$
Where C is any constant and the function $$f\left(x\right)$$ is called the integrand.

Note:
We can derive the exponential function $$x^n$$ as follows
Let the $$z=x^n$$
Differentiate with respect to x
$$\begin{gathered} dz=n{x}^{n-1}dx \\ dx={\displaystyle \frac{1}{n{z}^{n-1}}}dz \end{gathered}$$
We can change the denominator function as a numerator. So, it changed to negative exponential.
$$\begin{gathered} dx={\displaystyle \frac{1}{n}}{z}^{-(1-n)}dz \\ dx={\displaystyle \frac{1}{n}}{z}^{n-1}dz\Rightarrow {\displaystyle \frac{1}{n}}{z}^{{\displaystyle \frac{1}{n}}-1}dz \end{gathered}$$