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Question: How do you integrate \[\int {{x^5}{e^{ - {x^3}}}} \] with respect to \(x\) ?...

How do you integrate ∫x5e−x3\int {{x^5}{e^{ - {x^3}}}} with respect to xx ?

Explanation

Solution

The given question requires us to integrate a function of xx with respect to xx. We evaluate the given integral using the substitution method. We substitute the power of the exponential function as a new variable and then convert the integral in respect to the new variable.

Complete step by step answer:
The given question requires us to integrate a function x5e−x3{x^5}{e^{ - {x^3}}} in variable x.
So, we have, ∫x5e−x3dx\int {{x^5}{e^{ - {x^3}}}} dx
But, it is very difficult to integrate the function directly. So, we can assign a new variable to the power of the exponential function.
So, let −x3=t - {x^3} = t.
Then, differentiating both sides of the equation, we get,
⇒−3x2dx=dt\Rightarrow - 3{x^2}dx = dt
Finding the value of dxdx from the equation, we get,
⇒dx=dt−3x2\Rightarrow dx = \dfrac{{dt}}{{ - 3{x^2}}}
So, the integral ∫x5e−x3dx\int {{x^5}{e^{ - {x^3}}}} dx can be simplified by substituting the value of (dx)\left( {dx} \right) in terms of (dt)\left( {dt} \right) as obtained above. So, we get,
∫x5e−x3dx=∫x5e−x3(dt−3x2)\int {{x^5}{e^{ - {x^3}}}} dx = \int {{x^5}{e^{ - {x^3}}}} \left( {\dfrac{{dt}}{{ - 3{x^2}}}} \right)

Cancelling the common factors in numerator and denominator, we get,
⇒∫x3e−x3(dt−3)\Rightarrow \int {{x^3}{e^{ - {x^3}}}} \left( {\dfrac{{dt}}{{ - 3}}} \right)
Now, substituting the value of (−x3)\left( { - {x^3}} \right) as t, we get,
⇒∫(−t)et(dt−3)\Rightarrow \int {\left( { - t} \right){e^t}} \left( {\dfrac{{dt}}{{ - 3}}} \right)
Taking the constant out of the integral, we get,
⇒13∫tet(dt)\Rightarrow \dfrac{1}{3}\int {t{e^t}} \left( {dt} \right)
Now, we can evaluate the above integral using integration by parts method.
In integration by parts method, we integrate a function which is a product of two functions using a formula:
∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)∫g(x)dx]dx\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }

Hence, using integration by parts method and considering tt as first function and et{e^t} as the second function, we get,
⇒13[t∫etdt−∫[d[t]dt×∫etdt]dt]\Rightarrow \dfrac{1}{3}\left[ {t\int {{e^t}dt - \int {\left[ {\dfrac{{d\left[ t \right]}}{{dt}} \times \int {{e^t}dt} } \right]dt} } } \right]
Now, we know that the integral of ex{e^x} is ex{e^x}. Also, the derivative of x with respect to s is x. So, we get,
⇒13[tet−∫[1×et]dt]\Rightarrow \dfrac{1}{3}\left[ {t{e^t} - \int {\left[ {1 \times {e^t}} \right]dt} } \right]
⇒13[tet−∫etdt]\Rightarrow \dfrac{1}{3}\left[ {t{e^t} - \int {{e^t}dt} } \right]
⇒13[tet−et]+c\Rightarrow \dfrac{1}{3}\left[ {t{e^t} - {e^t}} \right] + c, where c is any arbitrary constant.
Taking et{e^t} common from all the terms, we get,
⇒et3[t−1]+c\Rightarrow \dfrac{{{e^t}}}{3}\left[ {t - 1} \right] + c
Now, substituting the value of t back into the expression, we get,
∴e−x33[−x3−1]+c\therefore \dfrac{{{e^{ - {x^3}}}}}{3}\left[ { - {x^3} - 1} \right] + c

So, e−x33[−x3−1]+c\dfrac{{{e^{ - {x^3}}}}}{3}\left[ { - {x^3} - 1} \right] + c is the solution of the integral ∫x5e−x3\int {{x^5}{e^{ - {x^3}}}} , where cc is any arbitrary constant.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant.