Power Rule
\(n\neq -1\)
Add one to the exponent, then divide by the new exponent.
Calculus Practice
Practice antiderivatives, indefinite and definite integrals, the Fundamental Theorem of Calculus, \(u\)-substitution, trig integrals, and integral properties.
Try each problem first, then click Show solution to check your work and see the complete step-by-step solution.
Keep These Rules Nearby
\(n\neq -1\)
Add one to the exponent, then divide by the new exponent.
Find an antiderivative, then evaluate upper minus lower.
Use substitution to reverse the chain rule.
Level 1
Start with simple antiderivatives, constants, and the integration power rule.
Problem 1
AntiderivativeStep-by-Step Solution
We need a function whose derivative is \(6x\).
Problem 2
Constant RuleStep-by-Step Solution
The antiderivative of a constant \(c\) is
\[ cx+C. \]Problem 3
Power RuleStep-by-Step Solution
Add one to the exponent:
\[ 5+1=6. \]Divide by the new exponent:
Level 2
Work with negative and fractional powers, polynomial integrals, and the special reciprocal rule.
Problem 4
PolynomialStep-by-Step Solution
Integrate each term separately.
Problem 5
Negative ExponentStep-by-Step Solution
Apply the power rule.
Rewrite using positive exponents:
\[ -\frac12x^{-2} = -\frac{1}{2x^2}. \]Problem 6
RadicalStep-by-Step Solution
Rewrite the radical:
\[ \sqrt{x} = x^{1/2}. \]Apply the power rule.
Problem 7
Reciprocal RuleStep-by-Step Solution
The power rule does not apply when the exponent is \(-1\).
Use the special reciprocal rule:
Level 3
Evaluate definite integrals and connect integration with differentiation.
Problem 8
Definite IntegralStep-by-Step Solution
An antiderivative of \(2x\) is
\[ x^2. \]Evaluate upper minus lower:
Problem 9
Fundamental TheoremStep-by-Step Solution
By the Fundamental Theorem of Calculus,
\[ \frac{d}{dx} \left[ \int_a^x f(t)\,dt \right] = f(x). \]Replace \(t\) by \(x\) in the integrand.
Problem 10
FTC + Chain RuleStep-by-Step Solution
Evaluate the integrand at the upper bound:
\[ \cos(x^2). \]Then multiply by the derivative of \(x^2\).
\[ \frac{d}{dx}(x^2) = 2x. \]Level 4
Apply basic trigonometric antiderivatives and reverse the chain rule with substitution.
Problem 11
Trig IntegralStep-by-Step Solution
Recall:
\[ \int \cos x\,dx = \sin x \] \[ \int \sin x\,dx = -\cos x. \]Problem 12
\(u\)-SubstitutionStep-by-Step Solution
Let
\[ u=x^2+1. \]Then
\[ du=2x\,dx. \]Rewrite:
Substitute back:
Problem 13
Constant FactorStep-by-Step Solution
Let
\[ u=x^2+4. \]Then
\[ du=2x\,dx. \]Therefore,
\[ x\,dx = \frac12du. \]Problem 14
Definite \(u\)-SubstitutionStep-by-Step Solution
Let
\[ u=x^2+1, \qquad du=2x\,dx. \]Convert the bounds:
\[ x=0 \Rightarrow u=1 \] \[ x=1 \Rightarrow u=2. \]Rewrite:
Level 5
Finish with integral properties, symmetry, signed area, and mixed concepts.
Problem 15
Integral PropertiesStep-by-Step Solution
Reversing the bounds changes the sign of a definite integral.
Problem 16
SymmetryStep-by-Step Solution
For an odd function,
\[ f(-x)=-f(x). \]Over an interval symmetric about zero, the positive and negative signed areas cancel.
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