The cube root of x is a number whose cube is x, in other words, (cube root of x)^3 = x. In this post, we will evaluate the integration of the cube root of x. To do so, we will use the power rule of derivatives which is given below:
$\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$ $ \cdots (i)$
where $c$ is an integration constant.
Also Read: Integration of root x
What is the integration of cube root of x? Let’s find out.
Integration of Cube Root of x
Question: Find the integration of cube root x, that is, find
$\int \sqrt[3]{x} dx$
Answer:
At first, by the rule of indices, we can write the cube root of $x$ as $x^{1/3}$. That is,
$\sqrt[3]{x}=x^{1/3}$
Using this fact, we have
$\int \sqrt[3]{x} dx$ $= \int x^{1/3} dx$
$=\dfrac{x^{1/3+1}}{1/3+1}+c \quad$ by the above power rule (i) of derivatives.
$=\dfrac{x^{1/3+1}}{4/3}+c$
$=\dfrac{3}{4} x^{1+1/3}+c$
$=\dfrac{3}{4} x \cdot x^{1/3}+c$
$=\dfrac{3}{4} x \sqrt[3]{x}+c$
So the integration of cube root of x is equal to $\dfrac{3}{4} x$ ∛x. In other words,
$\int$ ∛x $dx=\dfrac{3}{4} x$ ∛x +c where $c$ is an integration constant.
Also Read:
Integration of x/(1+x2) | Integration of x/(1+x) |
∫√(1+sin2x) dx | Integral of e^(x^2) |
Integration of log(sinx) | ∫xndx Formula, Proof |
Integration of 1 by Cube Root x
Question: Find the integration of 1 by cube root x, that is, calculate
$\int \dfrac{1}{\sqrt[3]{x}} dx$
Answer:
By the power rule of indices, we have 1/∛x $=1/x^{1/3}$. This can be further written as $x^{-1/3}$. Therefore, we have
$\int \dfrac{1}{\sqrt[3]{x}} dx$ $=\int x^{-1/3} dx$
Now applying the above power rule (i) of derivatives for $n=-1/3$, this is equal to
$= \int x^{-1/3} dx$
$=\dfrac{x^{-1/3+1}}{-1/3+1}+c$
$=\dfrac{x^{2/3}}{2/3}+c$
$=\dfrac{3}{2} x^{3/2}+c$
So the integration of 1 by cube root x is 3/2 x^{3/2}+c, where $c$ is an integration constant.
Definite integration of Cube Root x
Now, we will find a definite integral involving the cube root of x.
Question: Evaluate the integration of cube root x from 0 to 1.
$\int_0^1 \sqrt[3]{x} dx$
Answer:
From the above we know that $\int \sqrt[3]{x} dx=\dfrac{3}{4} x \sqrt[3]{x}$.
So we have
$\int_0^1 \sqrt[3]{x} dx$ =$[\dfrac{3}{4} x \sqrt[3]{x}]_0^1$
$=\dfrac{3}{4}[x \sqrt[3]{x}]_0^1$
$=\dfrac{3}{4}[1-0]$
$=\dfrac{3}{4}$.
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FAQs
Q1: What is the integration of cube root x?
Answer: The integration of cube root x is 3/4 x∛x +c.