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The integral of $x^{1}.x^{2}$ is basically the integration of $x^{3}$ and the integral of $x^{3}$ is $\dfrac{x^{4}}{4} + c$, where the “c” is a constant. The integral of $x^{3}$ is mathematically written as $\int x^{3}$. Integration is basically taking the antiderivative of a function, so in this case, we are taking the antiderivative of $x^{3}$.

In this topic, we will study how we can calculate the integral of $x^{1}.x^{2}$ by using several different methods of integration. We will also discuss some solved numerical examples for a better understanding of this topic.

## What Is Meant by the Integral of x^1.x^2?

The integral of $x^{1}.x^{2}$ or $x^{3}$ is taking the integration of function $x^{3}$ and the integration of $x^{3}$ is $\dfrac{x^{4}}{4} + c$. The integral of any function is basically a calculation of the area under the curve of the said function, so in this case, we calculate the area under the curve of the function $x^{3}$.

### Verifying Integral of x^1.x^2 Through Differentiation

We know that when we are calculating the integral of the function, then we are basically computing the antiderivative of the said function, so in this case, we need to find the function whose derivative is $x^{3}$. Let us calculate the derivative for $\dfrac{x^{4}}{4} + c$.

We can calculate the derivative by using the power rule of differentiation.

$\dfrac{d}{dx} x^{n} = n.x^{n-1}$

$\dfrac{d}{dx} \dfrac{x^{4}}{4} + c = 4 \times$ $\dfrac{x^{3}}{4} + 0 = x^{3}$

As we can see, the derivative of $\dfrac{x^{4}}{4} + c$ is $x^{3}$, so we have proved that antiderivative of $x^{3}$ is $\dfrac{x^{4}}{4} + c$.

### Formula for Integral of x^1.x^2

The formula for integral of $x^{1}.x^{2}$ or $x^{3}$ is given as:

$\int x^{1}.x^{2} dx = \int x^{3} dx = \dfrac{x^{4}}{4} + c$

Here:

$\int$ is the sign of integration

“c” is a constant

The expression dx shows that integration is done with respect to variable “x.”

__Proof__

We know that the integral for $x^{3}$ is $\dfrac{x^{4}}{4} + c$, and we can easily prove that by using the power rule of integration. According to the power rule of integration:

$\int x^{n} = \dfrac{x^{n+1}}{n+1} + c$

So, applying this to our function $x^{3}$:

$\int x^{3} = \dfrac{x^{4}}{4} + c$

Hence, we have proved the integration of $x^{1}. x^{2} = x^{3}$ is $\dfrac{x^{4}}{4} + c$.

### Integration of x^1.x^2 Using Integration by Parts

We can also verify the integral of $x^{3}$ by using the integration by parts method. The general formula for integration by parts can be written as:

$\int f(x). h(x) dx = f(x) \int h(x) – int [f^{‘}(x) \int h(x) dx] dx$

So when calculating the integral of $x^{3}$, $f(x) = x^{3}$ while $h(x) = 1$:

$\int x^{3} dx = \int x^{3}.1 dx$

$\int x^{3} dx = x^{3} \int 1 dx – \int [\frac{d x^{3}}{dx} \times \int 1dx] dx$

$\int x^{3} dx = x^{3}.x – \int [3x^{2}. x] dx + c$

$\int x^{3} dx = x^{3}.x – 3\int [x^{2}. x] dx + c$

$\int x^{3} dx = x^{3}.x – 3\int [x^{3}. dx + c$

$\int x^{3} dx + 3\int x^{3}. dx = x^{4} + c$

$4\int x^{3} dx = x^{4} + c$

$\int x^{3} dx = \dfrac{x^{4}}{4} + c$

Hence, we have proved the integration of $x^{1}. x^{2} = x^{3}$ is $\dfrac{x^{4}}{4} + c$.

### Definite Integral of x^1.x^2

The definite integral of $x^{1}.x^{2}$ is $\dfrac{b^{4}}{4} – \dfrac{a^{4}}{4}$, where a and b are lower and upper limits, respectively. So far, we have discussed indefinite integrals which are without any limits, so let us calculate if the integral has upper and lower limits for $x^{3}$.

Suppose we are given the upper and lower limits as “b” and “a” respectively for the function $x^{3}$, then the integration of $x. x^{2}$ will be:

$\int_{a}^{b} x^{3} = [\dfrac{x^{4}}{4}+ c ]_{a}^{b}$

$\int_{a}^{b} x^{3} = ( \dfrac{b^{4}}{4} + c) – ( \dfrac{a^{4}}{4} + c)$

$\int_{a}^{b} x^{3} = \dfrac{b^{4}}{4} + c – c – \dfrac{a^{4}}{4}$

$\int_{a}^{b} x^{3} = \dfrac{b^{4}}{4} – \dfrac{a^{4}}{4}$

Hence, we have proved that if the function $x^{3}$ has upper and lower limits of “b” and “a”, then the result is $\dfrac{b^{4}}{4} – \dfrac{a^{4}}{4}$.

**Example 1:** Evaluate the integral $x^{3}.e^{x}$.

__Solution:__

We can solve this function by using integration by parts. Let us take $x^{3}$ as the first function and $e^{x}$ as the second function. Then by definition of integral by parts, we can write the function as:

$\int x^{3}.e^{x} = x^{3} \int e^{x} dx – \int [\frac{d x^{3}}{dx} \times \int e^{x}dx] dx$

$\int x^{3}.e^{x} = x^{3}.e^{x} – \int [3x^{2}. e^{x}] dx$

$\int x^{3}.e^{x} = x^{3}. e^{x} – 3\int [x^{2}].e^{x} dx$

$\int x^{3}.e^{x} = x^{3}. e^{x} – 3\int [x^{2}e^{x}]. dx$

$\int x^{3}.e^{x} = x^{3}. e^{x} – 3I$

Let suppose $I = \int [x^{2}e^{x}] dx$

$I = x^{2} \int e^{x} dx – \int [\frac{d x^{2}}{dx} \times \int e^{x}dx] dx$

$I = x^{2}.e^{x} – \int [2x. e^{x}] dx$

$I = x^{2}. e^{x} – 2\int [x^.e^{x} dx$

$I = x^{2}. e^{x} – 2[e^{x}(x-1)]$

$I = e^{x}(x^{2}-2x + 2) + c$

Now putting this value back in the equation:

$\int x^{3}.e^{x} = x^{3}. e^{x} – 3 e^{x}(x^{2}-2x + 2) + c$

$\int x^{3}.e^{x} =e^{2}[ x^{3} – 3 (x^{2}-2x + 2)] + c$

**Example 3: **Evaluate the integral $x^{3}$ with upper and lower limits as $1$ and $0$, respectively.

__Solution:__

$\int_{0}^{1} x^{3} = [\dfrac{x^{4}}{4}+ c ]_{0}{1}$

$\int_{0}^{1} x^{3} = (\dfrac{(1)^{4}}{4} ) – ( \dfrac{(0)^{4}}{4} )$

$\int_{0}^{1} x^{3} = \dfrac{1}{4}$

**Practice Questions:**

- Evaluate the integral $\int \dfrac{x^{3}}{x.(x^{2}+1)}$.
- Evaluate integral of $2+1 x^{2}$.
- What is the integral of $x^{2}$?
- Evaluate the integral of x/(1+x^2).

Answer Keys:

1).

$\int \dfrac{x^{3}}{x.(x^{2}+1)} = \int \dfrac{x^{2}}{(x^{2}+1)}$

Subtracting and Adding the numerator expression by “1.”

$\int x^{3} dx = \int \dfrac{x^{2} + 1 – 1}{(x^{2}+1)}$

$\int x^{3} dx = \int \dfrac{(x^{2}+1)}{ (x^{2}+1)} dx – \int \dfrac{(1)}{ (x^{2}+1)} dx$

$\int x^{3} dx = \int 1 dx – \int \dfrac{(1)}{ (x^{2}+1)} dx$

$\int x^{3} dx = x – tan^{-1}x + c$

2).

We have to basically evaluate the integral of $3.x^{2}$.

$\int 3. x^{2} dx = 3 \int x^{2} dx$

$\int 3. x^{2} dx = 3 \dfrac{x^{3}}{3} + c$

$\int 3. x^{2} dx = \dfrac{x^{3}}{3} + c$

So the integral of $3.x^{2}$ is $\dfrac{x^{3}}{3} + c$.

3).

The integral of $x^{2}$ by using the power rule of integration will be:

$\int x^{2} dx = \dfrac{x^{2+1}}{2+1} + c = \dfrac{x^{3}}{3} + c$

4).

We will solve for the integral of $\dfrac{x}{1+x^{2}}$ by using the substitution method.

Let $u = 1 + x^{2}$

Taking derivatives on both sides.

$du = 0 + 2x dx$

$x.dx = \dfrac{du}{2}$

$\int \dfrac{x}{1+x^{2}} = \dfrac{1}{2} \int \dfrac{1}{u} dx$

$\int \dfrac{x}{1+x^{2}} = \dfrac{1}{2} ln|u| + c =\dfrac{1}{2} ln|1+x^{2}| + c$