int(x-a)(x^{n-1}+x^{n-2}a+ldots+a^{n-1})dx= ( | vedclass

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Question

(A) We know that the algebraic identity for the difference of two $n$-th powers is given by: $(x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1}) = x^n - a^n$. Sub

Vedclass · Accepted Answer

$\frac{x^{n+1}}{n+1}-a^n x+C$

Vedclass · Answer

(A) We know that the algebraic identity for the difference of two $n$-th powers is given by: $(x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1}) = x^n - a^n$. Substituting this into the integral,we get: $\int(x^n - a^n)dx$. Integrating term by term with respect to $x$: $\int x^n dx - \int a^n dx = \frac{x^{n+1}}{n+1} - a^n x + C$.

$\int(x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1})dx=$ (where $C$ is a constant of integration)

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