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# polyint

Polynomial integration

q = polyint(p,k)
q = polyint(p)

## Description

example

q = polyint(p,k) returns the integral of the polynomial represented by the coefficients in p using a constant of integration k.

example

q = polyint(p) assumes a constant of integration k = 0.

## Examples

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Evaluate the definite integral

$I={\int }_{-1}^{3}\left(3{x}^{4}-4{x}^{2}+10x-25\right)dx.$

Create a vector to represent the polynomial integrand $3{x}^{4}-4{x}^{2}+10x-25$. The ${\mathit{x}}^{3}$ term is absent and thus has a coefficient of 0.

p = [3 0 -4 10 -25];

Use polyint to integrate the polynomial using a constant of integration equal to 0.

q = polyint(p)
q = 1×6

0.6000         0   -1.3333    5.0000  -25.0000         0

Find the value of the integral by evaluating q at the limits of integration.

a = -1;
b = 3;
I = diff(polyval(q,[a b]))
I = 49.0667

Evaluate

$I={\int }_{0}^{2}\left({x}^{5}-{x}^{3}+1\right)\left({x}^{2}+1\right)dx$

Create vectors to represent the polynomials $p\left(x\right)={x}^{5}-{x}^{3}+1$ and $v\left(x\right)={x}^{2}+1$.

p = [1 0 -1 0 0 1];
v = [1 0 1];

Multiply the polynomials and integrate the resulting expression using a constant of integration k = 3.

k = 3;
q = polyint(conv(p,v),k)
q = 1×9

0.1250         0         0         0   -0.2500    0.3333         0    1.0000    3.0000

Find the value of I by evaluating q at the limits of integration.

a = 0;
b = 2;
I = diff(polyval(q,[a b]))
I = 32.6667

## Input Arguments

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Polynomial coefficients, specified as a vector. For example, the vector [1 0 1] represents the polynomial ${x}^{2}+1$, and the vector [3.13 -2.21 5.99] represents the polynomial $3.13{x}^{2}-2.21x+5.99$.

Data Types: single | double
Complex Number Support: Yes

Constant of integration, specified as a numeric scalar.

Example: polyint([1 0 0],3)

Data Types: single | double
Complex Number Support: Yes

## Output Arguments

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Integrated polynomial coefficients, returned as a row vector. For more information, see Create and Evaluate Polynomials.