Question

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

Answer 1

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

$$\Rightarrow 3{x^2}dx = dz \\\ \Rightarrow dx = \dfrac{{dz}}{{3{x^2}}} \\\$$

Comparing the exponential function $z = {x^n}$ with $dx = \dfrac{1}{n}{z^{\dfrac{1}{n} - 1}}dz$, so we can write

$$dx = \dfrac{1}{3}{z^{1 - 3}}dz \\\ dx = \dfrac{1}{3}{z^{\dfrac{1}{3} - 1}}dz \\\$$

Since, $n = 3$
Now we can substitute the ‘dx’ and ‘z’ value into the given equation
\int {{e^{{x^3}}}dx} = \int {{e^z}\dfrac{{dz}}{{3{x^2}}}} $$$$\int {{e^{{x^3}}}dx} = \int {{e^z}\dfrac{{dz}}{{3{x^2}}}}
We take the integral limit as $0$to $\infty$, we get
$\int {{e^{{x^3}}}dx} = \int {{e^z}\dfrac{1}{3}{z^{\dfrac{1}{3} - 1}}dz}$
$\int {{e^{{x^3}}}dx} = \int\limits_0^\infty {{e^z}\dfrac{1}{3}{z^{\dfrac{1}{3} - 1}}dz}$
Expand the integral values on the exponential function, we can get
$\int {{e^{{x^3}}}dx} = \dfrac{1}{3}\left( {\int\limits_0^\infty {{e^z}{z^{\dfrac{1}{3} - 1}}dz - \int\limits_z^\infty {{e^z}{z^{\dfrac{1}{3} - 1}}dz} } } \right) + c$
On compare the formula for indefinite integral $\Gamma (n,z) + d$ with the above derivative equation, the
$\int {{e^{{x^3}}}dx} = \dfrac{1}{3}\Gamma (\dfrac{1}{3},{x^3}) + d$
Where $d$ and $c$ are constant.
Hence, the integral of ${e^{{x^3}}}$is $\dfrac{1}{3}\Gamma (\dfrac{1}{3},{x^3}) + d$.

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:
$\int\limits_a^b {f(x)dx}$ Indefinite Integral:
Indefinite integrals are defined without upper and lower limits. It is represented as:
$\smallint f(x)dx = F(x) + C$
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

$$dz = n{x^{n - 1}}dx \\\ dx = \dfrac{1}{{n{z^{n - 1}}}}dz \\\$$

We can change the denominator function as a numerator. So, it changed to negative exponential.

$$dx = \dfrac{1}{n}{z^{ - (1 - n)}}dz \\\ dx = \dfrac{1}{n}{z^{n - 1}}dz \Rightarrow \dfrac{1}{n}{z^{\dfrac{1}{n} - 1}}dz \\\$$