integral of power
Integral of power
Primitive of power
Let \[n\in\mathbb{R},n\ne-1\], then we propose that \[\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\], where \[C\] is an arbitrary constant.
To prove that \[\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\], we use the power rule for derivatives to show that, \[\frac{d}{dx}\left( \frac{x^{n+1}}{n+1}+C \right)=(n+1)\left( \frac{x^{(n+1)-1}}{n+1} \right)=x^{n}\] as \[\frac{1}{n+1}\] is a constant.
Since the first fundamental theorem of calculus tells us that if \[F(x)\] is an antiderivative of \[f(x)\], then \[F^{\prime}(x)=f(x)\] must hold true. We proposed \[F(x)=\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\] and shown that \[F^{\prime}(x)=x^{n}=f(x)\] holds true. Therefore, by the first fundamental theorem of calculus, we have proven that \[\frac{x^{n+1}}{n+1}+C\] is indeed an antiderivative of \[x^{n}\], thus establishing the validity of this formula.
\[\int \left( ax+b \right)^{n}\,dx\]
Based on what we've proven above, we can now prove \[\int \left( ax+b \right)^{n}\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+C\] where \[n\ne1\] with integration by substitution.
Let \[f(u)=u^{n}\], \[g(x)=ax+b\], then \[g^{\prime}(x)=a\]. Now based on this \[f(g(x))\cdot g^{\prime}(x)=\left( ax+b \right)^{n}\cdot a\], therefore to remove the \[a\], simple do \[\frac{1}{a}\int (ax+b)^{n}\cdot a\,dx\].
Then, \[F(u)=\int u^{n}\,du=\frac{u^{n+1}}{n+1}\]. Substituting everything back into \[\int f(g(x))\cdot g^{\prime}(x)\,dx=F(g(x))+C\],
Integral of power
For all \[n\in\mathbb{R},n\ne-1\], then \[\int_{0}^{b}x^{n}\,dx=\frac{b^{n+1}}{n+1}\].
From the second fundamental theorem of calculus, \[\int_{0}^{b}x^{n}\,dx=\biggl[ F(x) \biggr]_{0}^{b}=F(b)-F(0)\] where \[F(x)\] is an antiderivative of \[x^{n}\]. By the primitive of power, \[\frac{x^{n+1}}{n+1}\] is one of the possible antiderivatives of \[x^{n}\].
Then,
proving the formula.