Question

Find the indefinite integral and check your result by differentiation. $$ \int(x+3) d x $$

Short answer

The indefinite integral of \( (x + 3) \) is \( \frac{1}{2} x^2 + 3x + C \)

Step-by-step solution

Calculate the Integral

We will use the power rule for integration, which states that \( \int x^n dx = \frac{1}{n+1} x^{n+1} + C \), where \( C \) is the constant of integration. Our integral can be split into two separate integrals. \[ \int(x+3) dx = \int x dx + \int 3 dx. \] For \( \int x dx \), \( n = 1 \), so the result of this integration according to the power rule will be \( \frac{1}{1 + 1} x^{1 + 1} = \frac{1}{2} x^2 \). For \( \int 3 dx \), there's no \( x \) term, meaning it's equivalent to \( 3 \int dx \). The integral of \( dx \) is simply \( x \), so \( \int 3 dx = 3x \). Combining these results, our integral is \( \frac{1}{2} x^2 + 3x + C \).

Check the result by differentiation

In order to check our integration, we must differentiate our result and compare it to the original integrand \( x + 3 \). The derivative of \( \frac{1}{2} x^2 \) is \( x \) using the power rule and the derivative of \( 3x \) is \( 3 \). The derivative of the constant \( C \) is 0. Therefore, the derivative of \( \frac{1}{2} x^2 + 3x + C \) is \( x + 3 \), which is our original integrand, thus confirming our integral is correct.

Key concepts

Integration by Power Rule

The power rule for integration is a fundamental tool in calculus for finding the indefinite integral of functions that can be expressed as a power of a variable. When you're faced with an integral of the form \( \int x^n dx \), the power rule states that you can add one to the exponent and then divide by the new exponent, so the integral becomes \( \frac{1}{n+1} x^{n+1} \), provided that \( n \eq -1 \).

It's essential to apply the power rule term-wise to functions like \( x + 3 \), splitting the integral into \( \int x dx + \int 3 dx \). For \( x^1 \), the new exponent is 2, so the result is \( \frac{1}{2} x^2 \). Constants like 3 are treated differently - since the derivative of \( x \) is 1, the antiderivative of a constant \( a \) is \( ax \).

Remember, the power rule for integration is applicable only when the variable is raised to a real number power and is not valid for expressions where the variable is in the denominator with an exponent of -1 (as in the integral of \( 1/x \) or \(x^{-1}\)).

Constant of Integration

A crucial aspect of indefinite integrals is the constant of integration, often represented by \( C \). Since differentiation wipes out any constant (as its derivative is 0), when we integrate, we must acknowledge the possibility of such a constant. For any function \( F(x) \), its derivative is \( f(x) \) ; therefore, the antiderivative or integral of \( f(x) \) must be \( F(x) + C \), where \( C \) represents an unknown constant that could be any real number.

For example, when integrating \( x + 3 \), the result includes a term \( + C \), written as \( \frac{1}{2} x^2 + 3x + C \). This acknowledgment of the constant is essential because, without it, we would miss an entire family of functions that are vertically shifted by a constant amount from our basic antiderivative.

Differentiation to Check Integration

After integrating a function, it is prudent to check your work. One reliable way to verify an indefinite integral is by differentiating the resulting expression and comparing it with the original integrand. If they match, you have likely integrated correctly.

The process is straightforward: you differentiate each term of the antiderivative \( \frac{1}{2} x^2 + 3x + C \) using the familiar power rule in reverse. The derivative of \( x^2 \) is \( 2x \) and when multiplied by the 1/2 coefficient, it simplifies to \( x \). Likewise, the derivative of \( 3x \) is 3, and the derivative of the constant \( C \) is zero, because constants do not change and hence have no rate of change. Therefore, if the differentiation of your integrated function returns the original function you started with, such as \( x + 3 \) in this case, then you have successfully found the indefinite integral of the function.