Q. \[ \int x^4\,dx \]

Q: How do you use the power rule for \int x^n\,dx ?

If n \neq -1, then \int x^n\,dx = \frac{x^{n+1}}{n+1} + C . For n=4: \frac{x^5}{5} + C

Q: What is the derivative of \frac{x^5}{5}+C to check?

\frac{d}{dx}\left(\frac{x^5}{5}\right)=\frac{5x^4}{5}=x^4. Also \frac{d}{dx}(C)=0

Q: What is the definite integral \int_{-1}^{1} x^4\,dx ?

x^4 is even, so \int_{-1}^{1} x^4\,dx = 2\int_{0}^{1} x^4\,dx = 2\left[\frac{x^5}{5}\right]_{0}^{1}=\frac{2}{5}

Q: What if the integrand were x^{-2}? Does the rule still work?

The same power rule works if n\neq -1. For x^{-2}: \int x^{-2}\,dx = \frac{x^{-1}}{-1}+C = -x^{-1}+C

Answer

To find \( \int x^4\,dx \), use the power rule: \( \int x^n\,dx = \frac{x^{n+1}}{n+1}+C \) for \( n\neq -1 \).

Here \( n=4 \), so \( n+1=5 \):

\[
\int x^4\,dx=\frac{x^5}{5}+C
\]

Detailed Explanation

We want to find the indefinite integral:

\[
\int x^4 \, dx
\]

Step 1: Identify the power rule for integrals.

For any integer power \(n \neq -1\), the rule is:

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

Here, we match \(x^4\) to the form \(x^n\). That means \(n = 4\).

Step 2: Apply the power rule.

Substitute \(n = 4\) into the formula:

\[
\int x^4 \, dx = \frac{x^{4+1}}{4+1} + C
\]

Step 3: Simplify exponents and the denominator.

\(4+1 = 5\), so \(x^{4+1} = x^5\). Also \(4+1 = 5\) in the denominator:

\[
\int x^4 \, dx = \frac{x^5}{5} + C
\]

Final Answer:

\[
\int x^4 \, dx = \frac{x^5}{5} + C
\]

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Calculus FAQ

What is \(\int x^4\,dx\) ?

\( \int x^4\,dx = \frac{x^5}{5} + C \)

How do you use the power rule for \(\int x^n\,dx\) ?

If \(n \neq -1\), then \( \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \). For \(n=4\): \( \frac{x^5}{5} + C \)

What is the derivative of \(\frac{x^5}{5}+C\) to check?

\(\frac{d}{dx}\left(\frac{x^5}{5}\right)=\frac{5x^4}{5}=x^4\). Also \(\frac{d}{dx}(C)=0\)

What is the definite integral \(\int_{0}^{2} x^4\,dx\) ?

\( \int_{0}^{2} x^4\,dx = \left[\frac{x^5}{5}\right]_{0}^{2} = \frac{32}{5}\)

What is the definite integral \(\int_{-1}^{1} x^4\,dx\) ?

\(x^4\) is even, so \( \int_{-1}^{1} x^4\,dx = 2\int_{0}^{1} x^4\,dx = 2\left[\frac{x^5}{5}\right]_{0}^{1}=\frac{2}{5}\)

What if the integrand were \(x^{-2}\)? Does the rule still work?

The same power rule works if \(n\neq -1\). For \(x^{-2}\): \( \int x^{-2}\,dx = \frac{x^{-1}}{-1}+C = -x^{-1}+C\)

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