Mathematics · Calculus
Definite Power Integral Calculator
Evaluate the definite integral of axⁿ between lower and upper bounds.
Inputs and results stay in this browser. Change one value at a time to explore the relationship.
Calculation steps
- Antiderivative coefficient = 3 ÷ 3 = 1.
- Evaluate bounds: F(2) = 8; F(0) = 0.
- Integral = 8 − 0 = 8.
Understand Definite power integral
One idea, three depths
Choose how deeply to explain Definite power integral
Definite power integral: Evaluate the definite integral of axⁿ between lower and upper bounds.
Age 5Explain it to a 5-year-oldStart with a picture
Imagine using Definite power integral to answer this question: evaluate the definite integral of axⁿ between lower and upper bounds? Enter Coefficient a, Power n, Lower bound, and 1 other input; the calculator shows Definite integral. For example: The integral of 3x² from 0 to 2 is [x³]₀² = 8. The answer tells you Definite integral.
Age 15Explain it to a 15-year-oldConnect it to the formula
The Fundamental Theorem of Calculus subtracts the antiderivative at the lower bound from its value at the upper bound. The rule is ∫[u,v] axⁿ dx = a/(n+1)(vⁿ⁺¹ − uⁿ⁺¹), n ≠ −1. Its input values are Coefficient a, Power n, Lower bound, Upper bound, and the main result is Definite integral. For example: The integral of 3x² from 0 to 2 is [x³]₀² = 8.
CollegeExplain it at college levelState the model precisely
This calculator evaluates the stated definite power integral relation over the valid real-number domain stated below. The implemented relation is ∫[u,v] axⁿ dx = a/(n+1)(vⁿ⁺¹ − uⁿ⁺¹), n ≠ −1, evaluated from Coefficient a, Power n, Lower bound, Upper bound to produce Definite integral. The Fundamental Theorem of Calculus subtracts the antiderivative at the lower bound from its value at the upper bound. Evaluate both bounds and subtract lower from upper; a negative result can be valid signed area.
Inputs and valid domain
- Coefficient a must be a finite real number.
- Power n must be a finite real number.
- Lower bound must be a finite real number.
- Upper bound must be a finite real number.
Important boundary: Evaluate both bounds and subtract lower from upper; a negative result can be valid signed area.
The formula
∫[u,v] axⁿ dx = a/(n+1)(vⁿ⁺¹ − uⁿ⁺¹), n ≠ −1
How the calculator works through it
It substitutes Coefficient a, Power n, Lower bound, Upper bound into the formula and exposes every numerical step above. The main output is Definite integral, accompanied by Upper antiderivative value, Lower antiderivative value.
Read the result correctly
The Definite integral is the direct answer to “evaluate the definite integral of axⁿ between lower and upper bounds.” Read it with the units shown beside the inputs; a sign, angle, percentage or rate changes what the number means.
A worked check
The integral of 3x² from 0 to 2 is [x³]₀² = 8.
Where this model stops being reliable
Evaluate both bounds and subtract lower from upper; a negative result can be valid signed area.
Learn it by changing one value
Begin with the worked example, then change one value while keeping the others fixed. Compare the new result and calculation steps to identify which part of the formula changed.
Dictionary terms behind this calculator
Before studying the codeWhat you should know firstUse the calculator immediately, or check the foundations before reading the implementation.
These foundations help you understand why Definite power integral works. They never block the calculator, and “optional” means useful context rather than a hidden requirement.
Hard requirements
- Reading formulas and substituting values
Definite power integral uses ∫[u,v] axⁿ dx = a/(n+1)(vⁿ⁺¹ − uⁿ⁺¹), n ≠ −1. You need to recognise what each side represents before substituting the stated inputs or rearranging the relationship.
Review this foundation about 4 min
Strong support
- Derivatives as rates of change
Rates of change explain the local behaviour captured or approximated by Definite power integral.
Review this foundation about 7 min
Optional enrichment
- Accumulation and integral notation
Integral notation connects Definite power integral to accumulated change, area and continuous totals.
Review this foundation about 6 min
Mathematics → algorithm → program
Implement this calculation in code
These are direct reference implementations of the calculator's principal relationship and first output. They run locally and include a small known-answer check where the language supports it.
Algorithm
- Read Coefficient a, Power n, Lower bound, Upper bound.
- Evaluate the principal relationship: ∫[u,v] axⁿ dx = a/(n+1)(vⁿ⁺¹ − uⁿ⁺¹), n ≠ −1.
- Return Definite integral and check the domain conditions described above.
Python
from math import *
def definite_power_integral(a, n, x1, x2) -> float:
return ((a / (n + 1.0)) * (pow(x2, (n + 1.0)) - pow(x1, (n + 1.0))))
assert abs(definite_power_integral(3, 2, 0, 2) - 8) < 1e-6 * max(1.0, abs(8))
C
#include <assert.h>
#include <math.h>
double definite_power_integral(double a, double n, double x1, double x2) {
return ((a / (n + 1.0)) * (pow(x2, (n + 1.0)) - pow(x1, (n + 1.0))));
}
int main(void) {
const double expected = 8;
const double actual = definite_power_integral(3, 2, 0, 2);
assert(fabs(actual - expected) < 1e-6 * fmax(1.0, fabs(expected)));
}
C++
#include <cassert>
#include <cmath>
#include <numbers>
double definite_power_integral(double a, double n, double x1, double x2) {
return ((a / (n + 1.0)) * (std::pow(x2, (n + 1.0)) - std::pow(x1, (n + 1.0))));
}
int main() {
constexpr double expected = 8;
const double actual = definite_power_integral(3, 2, 0, 2);
assert(std::fabs(actual - expected) < 1e-6 * std::fmax(1.0, std::fabs(expected)));
}
Linux x86-64 assembly
x86-64 NASM · System V ABI · Linux · SSE2 with libm where required
; double definite_power_integral(double a, double n, double x1, double x2)
; Linux x86-64 NASM · System V ABI · first eight doubles in xmm0–xmm7
extern pow
global definite_power_integral
section .text
definite_power_integral:
push rbp
mov rbp, rsp
sub rsp, 128
movsd [rbp-8], xmm0
movsd [rbp-16], xmm1
movsd [rbp-24], xmm2
movsd [rbp-32], xmm3
mov rax, 0x3ff0000000000000
movq xmm0, rax
movsd [rbp-64], xmm0
movsd xmm0, [rbp-16]
addsd xmm0, [rbp-64]
movsd [rbp-56], xmm0
movsd xmm0, [rbp-8]
divsd xmm0, [rbp-56]
movsd [rbp-48], xmm0
mov rax, 0x3ff0000000000000
movq xmm0, rax
movsd [rbp-96], xmm0
movsd xmm0, [rbp-16]
addsd xmm0, [rbp-96]
movsd [rbp-88], xmm0
movsd xmm0, [rbp-32]
movsd xmm1, [rbp-88]
call pow wrt ..plt
movsd [rbp-80], xmm0
mov rax, 0x3ff0000000000000
movq xmm0, rax
movsd [rbp-120], xmm0
movsd xmm0, [rbp-16]
addsd xmm0, [rbp-120]
movsd [rbp-112], xmm0
movsd xmm0, [rbp-24]
movsd xmm1, [rbp-112]
call pow wrt ..plt
movsd [rbp-104], xmm0
movsd xmm0, [rbp-80]
subsd xmm0, [rbp-104]
movsd [rbp-72], xmm0
movsd xmm0, [rbp-48]
mulsd xmm0, [rbp-72]
movsd [rbp-40], xmm0
movsd xmm0, [rbp-40]
leave
ret
MATLAB
function result = definite_power_integral(a, n, x1, x2)
result = ((a / (n + 1.0)) * ((x2 ^ (n + 1.0)) - (x1 ^ (n + 1.0))));
end
Wolfram Language
ClearAll[mwCalculate];
mwCalculate[a_, n_, x1_, x2_] := ((a / (n + 1.0)) * ((x2 ^ (n + 1.0)) - (x1 ^ (n + 1.0))));
Continue in mathematical software
The downloaded file includes your current inputs and first calculated result. It is created locally.
Floating-point answers can differ slightly by language, compiler and processor. Compare within a suitable tolerance rather than assuming every decimal representation will be identical.
Supporting sourcesAcademic referencesPrimary standards, textbooks and complete citations
Standards, reading and academic references
Use the calculator as the worked interaction, then consult the primary standards and academic textbooks listed below. MW SysArc links to the original sources; the explanation on this page is original and does not reproduce them.
Calculus Volume 1
Read OpenStax Calculus: Derivatives and integrationCite this book
- APA 7
- Strang, G., & Herman, E. (2016). Calculus volume 1. OpenStax. https://openstax.org/books/calculus-volume-1/pages/1-introduction
- MLA 9
- Strang, Gilbert, and Edwin Herman. Calculus Volume 1. OpenStax, 2016, https://openstax.org/books/calculus-volume-1/pages/1-introduction.
- Chicago author-date
- Strang, Gilbert, and Edwin Herman. 2016. Calculus Volume 1. Houston, TX: OpenStax. https://openstax.org/books/calculus-volume-1/pages/1-introduction.
OpenStax entries are free to read online. Follow the licence shown on each linked source before redistributing or adapting its content.
Reuse the page responsiblyCite this pageAPA, MLA, Chicago, Harvard, BibTeX and RIS
These formats cite this calculator page itself. They are separate from the academic references above, which support the mathematical method and terminology.
APA 7
MW SysArc. (2026, July 21). Definite Power Integral Calculator. MW SysArc Tools. https://math.mwsysarc.com/calculus/definite-power-integral
MLA 9
MW SysArc. “Definite Power Integral Calculator.” MW SysArc Tools, 21 July 2026, https://math.mwsysarc.com/calculus/definite-power-integral. Accessed 31 Aug. 2026.
Chicago 17
MW SysArc. “Definite Power Integral Calculator.” MW SysArc Tools. Published July 21, 2026. Accessed August 31, 2026. https://math.mwsysarc.com/calculus/definite-power-integral.
Harvard
MW SysArc (2026) ‘Definite Power Integral Calculator’, MW SysArc Tools. Published 21 July 2026. Available at: https://math.mwsysarc.com/calculus/definite-power-integral (Accessed: 31 August 2026).
BibTeX and RIS records
BibTeX
@misc{mwsysarc_definite_power_integral_2026,
author = {{MW SysArc}},
title = {Definite Power Integral Calculator},
howpublished = {MW SysArc Tools},
year = {2026},
url = {https://math.mwsysarc.com/calculus/definite-power-integral},
note = {Published July 21, 2026; accessed August 31, 2026}
}RIS
TY - ELEC
AU - MW SysArc
TI - Definite Power Integral Calculator
T2 - MW SysArc Tools
PY - 2026
DA - 2026-07-21
Y2 - 2026-08-31
UR - https://math.mwsysarc.com/calculus/definite-power-integral
N1 - Published July 21, 2026
ER -Clear answers
Frequently asked questions
What does the Definite power integral do?
Evaluate the definite integral of axⁿ between lower and upper bounds.
How does the Definite power integral work?
The calculator applies ∫[u,v] axⁿ dx = a/(n+1)(vⁿ⁺¹ − uⁿ⁺¹), n ≠ −1. The Fundamental Theorem of Calculus subtracts the antiderivative at the lower bound from its value at the upper bound.
What can I learn from the Definite power integral?
It connects the mathematical rule to your chosen numbers and shows each calculation step. Change one input at a time to see how the result responds.
Does MW SysArc receive or store what I enter?
No. The calculation runs locally in your browser. MW SysArc does not receive or store your calculation inputs.
How should I use the result?
Use the steps to understand the method, then verify important school or professional work using the notation and rounding rules required in your setting.
Last reviewed . Calculations tested .