What is integral of x^3/4 | ∫x34.dx=?

 Understanding the Integral of ( x^{3/4} )

Calculating the integral of a function is a fundamental concept in calculus which helps us find the area under a curve, among other applications. In this article, we will explore the integration of the function ( f(x) = x^{3/4} ).

Integral Definition

The integral of a function ( f(x) ) is denoted as:

$\int f(x) , dx$

For our specific case, we want to compute:

$\int x^{3/4} , dx$

Power Rule of Integration

To solve this integral, we apply the power rule of integration. The power rule states that for any real number ( n \neq -1 ):

$\int x^n , dx = \frac{x^{n+1}}{n+1} + C$

where ( C ) is the constant of integration.

Applying the Power Rule

In our case, ( n = \frac{3}{4} ). We first calculate ( n + 1 ):

$n + 1 = \frac{3}{4} + 1 = \frac{3}{4} + \frac{4}{4} = \frac{7}{4}$

Now we can substitute ( n ) into the integral formula:

$\int x^{3/4} , dx = \frac{x^{7/4}}{7/4} + C$

To simplify this expression, we multiply by the reciprocal of ( \frac{7}{4} ):

$\int x^{3/4} , dx = \frac{4}{7} x^{7/4} + C$

Final Result

Thus, the integral of ( x^{3/4} ) is:

$\int x^{3/4} , dx = \frac{4}{7} x^{7/4} + C$

where ( C ) represents the arbitrary constant of integration, which reflects that there are infinitely many antiderivatives for every function.

Applications

The integral we derived can be applied in various fields such as physics, engineering, and economics, particularly when dealing with problems related to area calculation, volume of solids, and rate of change over time.

Conclusion

Understanding the integral of basic power functions like ( x^{3/4} ) enhances our ability to tackle more complex problems in calculus. By practicing the power rule and familiarizing ourselves with integration techniques, we become more proficient in mathematical analysis and application.

In summary, the integral of ( x^{3/4} ) is:

$\int x^{3/4} , dx = \frac{4}{7} x^{7/4} + C$

This provides a solid basis for further exploration into integration and its various applications.

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x.e−x integral | What is integral of x.e^-x? | ∫x.e−x.dx=?

Understanding the Integral of ( x \cdot e^{-x} )

Calculating the integral ( \int x \cdot e^{-x} , dx ) may seem daunting at first glance, but with the application of integration techniques such as integration by parts, we can arrive at a solution. This integral combines polynomial and exponential functions, giving it unique properties that are useful in various fields, including physics and engineering.

Step-by-Step Solution

To evaluate ( \int x \cdot e^{-x} , dx ), we will use integration by parts, a method that leverages the product rule of differentiation. The formula for integration by parts is:

$\int u , dv = uv - \int v , du$

In this case, we can let:

( u = x ) → which implies ( du = dx )

( dv = e^{-x} , dx ) → which implies ( v = -e^{-x} )

Applying Integration by Parts

Substituting these choices into the integration by parts formula:

$\int x e^{-x} , dx = uv - \int v , du$

$= x(-e^{-x}) - \int (-e^{-x}) , dx$

$= -x e^{-x} + \int e^{-x} , dx$

Next, we need to compute ( \int e^{-x} , dx ):

$\int e^{-x} , dx = -e^{-x}$

Substituting this back into our previous result:

$\int x e^{-x} , dx = -x e^{-x} - e^{-x} + C$

Where ( C ) is the constant of integration.

Final Result

Combining the terms, we arrive at the final solution:

$\int x e^{-x} , dx = -e^{-x}(x + 1) + C$

Applications and Importance

The integral ( \int x e^{-x} , dx ) is particularly significant in probability theory, specifically in calculating moments of certain distributions like the exponential distribution. The presence of ( e^{-x} ) in the integral signifies damping or decay, which is observed in processes described by exponential functions.

Moreover, this integral serves as a standard example in calculus courses, demonstrating the application of integration by parts. Understanding how to tackle such integrals is crucial for students and professionals in mathematics, physics, engineering, and economics.

Conclusion

In summary, the integral ( \int x e^{-x} , dx ) can be elegantly solved using integration by parts, yielding the result ( -e^{-x}(x + 1) + C ). This example showcases not only a fundamental calculus technique but also the broader applicability of integrals involving exponential functions in various scientific fields.

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