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Integral Calculator

Computes indefinite and definite integrals of polynomial functions using the power rule for integration. For definite integrals, the user enters upper and lower limits and the calculator returns the exact area under the curve.

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Integral Calculator Logic

xndx=xn+1/(n+1)+C∫xⁿ dx = xⁿ⁺¹/(n+1) + C
Disclaimer: Results are estimates only. Always verify important calculations with a qualified professional before making decisions. Learn about our methodology.

Integration is the reverse of differentiation. The integral of f(x) is the function F(x) whose derivative equals f(x). For polynomial functions, the power rule for integration gives a direct method: the integral of axⁿ is axⁿ⁺¹/(n+1) + C, where C is the constant of integration. This calculator accepts any polynomial in x, returns the indefinite integral (antiderivative), and can also work out the definite integral between two limits to find the area under the curve. It covers the two core integration tasks that appear in most A-level and university calculus courses.

The Power Rule for Integration

The rule is: integral of xⁿ dx = xⁿ⁺¹/(n+1) + C. This is the reverse of the differentiation power rule. To figure out the integral of each term, increase the exponent by 1 and divide the coefficient by the new exponent. As a result, differentiating the integral always returns the original function, which provides a built-in check: if you differentiate F(x) and do not get f(x) back, an error occurred. Given that the rule is applied term by term for polynomials, integrating a long polynomial is just as direct as a short one. The Paul's Online Math Notes integration reference provides thorough worked examples covering the power rule and definite integrals at the calculus 1 level.

Term f(x)Integral F(x)Power rule step
x⁴/4 + CExponent 3+1=4, divide by 4
6x²2x³ + CExponent 2+1=3, 6/3 = 2
5x5x²/2 + CExponent 1+1=2, 5/2 = 2.5
44x + CConstant = 4x^0, exponent 0+1=1
3x² - 2x + 1x³ - x² + x + CSum rule: integrate each term

The Constant of Integration

The indefinite integral always includes +C, representing an unknown constant. This is because any constant added to F(x) has a derivative of zero, so F(x), F(x)+5, and F(x)-7 all differentiate to f(x). That said, when solving an initial value problem where you know F at a specific point, you substitute that condition to build up the specific value of C and narrow down the unique solution from the family of antiderivatives. For instance, if F(0) = 3 and F(x) = x² + C, then 0 + C = 3, so C = 3 and F(x) = x² + 3.

Definite Integrals and Area Under the Curve

A definite integral has specified lower and upper limits: integral from a to b of f(x) dx. By the Fundamental Theorem of Calculus, this equals F(b) - F(a), where F is any antiderivative of f. The result represents the signed area between the curve and the x-axis over the interval [a, b]. With that in mind, areas below the x-axis contribute negatively; the net area and the total area are different quantities when f(x) changes sign within the interval.

To find total area (always positive), you split the interval at the x-intercepts of f, carry out separate definite integrals over each sub-interval, and sum their absolute values. The Monash University Fundamental Theorem of Calculus guide explains the connection between antiderivatives and area clearly. On top of that, the CK-12 definite integral lesson provides interactive examples for both positive and negative area scenarios. Our Derivative Calculator is the companion for the reverse operation, and our Limit Calculator handles the limit definition of the definite integral when you want to understand the underlying foundation.

Integral Applications in Physics and Engineering

Integration is used to recover cumulative quantities from rates. If velocity is given as a polynomial v(t) = 6t² - 4t + 2, the integral gives displacement s(t) = 2t³ - 2t² + 2t + C. The constant C is determined by an initial condition such as s(0) = 5, which gives C = 5. With that in mind, this pattern of integrating a rate to build up a total quantity appears across physics, economics, and engineering. In economics, if a marginal cost function MC(q) is given as a polynomial in production quantity q, integrating gives the total cost function. On top of that, engineers figure out the area under stress-strain curves by integration to determine the energy absorbed by a material before failure. These practical applications demonstrate why the integral is one of the most useful operations in applied mathematics, and why learning to carry out integration confidently opens a wide range of quantitative analysis tasks. Given that all these examples use polynomial functions, this calculator directly supports the core computation in each case. As students figure out more advanced integration techniques such as integration by parts or trigonometric substitution, the power rule remains the foundational skill they always return to for polynomial terms within more complex expressions.

Founder's Real-World Experience
Muhammad Shahbaz Siddiqui

Muhammad Shahbaz Siddiqui

Founder, TheCalculatorsHub

How a Year 13 student used the definite integral to calculate the area under a velocity curve and verify a displacement answer

In April 2026, an A-Level Mathematics student in Cardiff was revising for her final exam when she encountered a question asking for the total displacement of an object between t = 0 and t = 4 seconds, given the velocity function v(t) = 3t² - 8t + 4. She had calculated the definite integral manually and obtained 16, but her revision guide showed the answer as 8 and she could not figure out where the discrepancy arose.

She entered 3x^2 - 8x + 4 into the Integral Calculator. The indefinite integral returned as x³ - 4x² + 4x + C. She then entered limits 0 and 4. The calculator returned the definite integral as 4³ - 4(4²) + 4(4) - 0 = 64 - 64 + 16 = 16. This matched her answer. She then checked whether the velocity function crossed zero in [0,4] by setting 3t² - 8t + 4 = 0, which factors as (3t - 2)(t - 2) = 0, giving roots at t = 2/3 and t = 2. Since the velocity is negative between t = 2/3 and t = 2, the net displacement (signed integral) is 16, but the total distance travelled (unsigned integral) requires splitting the integral at the roots.

The revision guide answer of 8 was asking for net signed displacement, not total distance. Her answer of 16 was correct for the unsigned (total distance) interpretation. Having the calculator confirm her anti-derivative and the definite integral value allowed her to narrow down the source of the discrepancy to a question-interpretation issue rather than a calculation error. She scored full marks on the integration section of her final exam in June 2026.

Anti-derivative x³-4x²+4x confirmed; definite integral from 0 to 4 = 16 matched student calculationDiscrepancy traced to net vs total displacement interpretation; revision guide answer (8) was for signed integral onlyFull marks on A-Level integration section June 2026; question-interpretation distinction now understood