Integrals

A faster exact method

An antiderivative reverses differentiation. If differentiating a function gives the curve you are integrating, evaluate that function at the bounds and subtract. This is the Fundamental Theorem of Calculus.

For the square curve, the power rule works in reverse. The result agrees with what the rectangles approach.

Where the basic integration rules come from

Every indefinite-integral rule can be checked by differentiating its answer. If $F'(x)=f(x)$, then $\int f(x)\,dx=F(x)+C$. The constant $C$ is needed because differentiating any constant gives zero.

1. 1. Constants, multiples, and sums

   Differentiation distributes across sums and constant multiples. It also turns $cx$ into the constant $c$. Reverse those three facts together:

   $$\begin{aligned}\frac{d}{dx}\bigl(AF(x)+BG(x)+cx\bigr)&=Af(x)+Bg(x)+c\\[4pt]\int\bigl(Af(x)+Bg(x)+c\bigr)\,dx&=AF(x)+BG(x)+cx+C\end{aligned}$$

   Here $F'=f$ and $G'=g$. This is why you can integrate a polynomial one term at a time.

2. 2. Powers

   The derivative power rule lowers an exponent by one. To reverse it, raise the exponent first, then divide by the new exponent:

   $$\begin{aligned}\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)&=\frac{(n+1)x^n}{n+1}=x^n\\[4pt]\int x^n\,dx&=\frac{x^{n+1}}{n+1}+C,\qquad n\ne-1\end{aligned}$$

   For example, $\int x^2\,dx$ becomes $x^3/3+C$. The rule cannot use $n=-1$ because that would divide by zero. Apply it on an interval where the power is defined.

3. 3. The reciprocal becomes a logarithm

   The missing power-rule case has its own antiderivative. On either side of zero, the derivative of the logarithm of absolute value is the reciprocal:

   $$\frac{d}{dx}\ln|x|=\frac{1}{x}\quad(x\ne0)\qquad\Longrightarrow\qquad\int\frac{1}{x}\,dx=\ln|x|+C$$

   Work on an interval that does not cross zero, where the integrand is defined.

4. 4. Exponentials

   The natural exponential differentiates to itself. For any other positive base, differentiation introduces a logarithm factor, so integration must divide by it:

   $$\begin{aligned}\frac{d}{dx}e^x&=e^x&\int e^x\,dx&=e^x+C\\[4pt]\frac{d}{dx}a^x&=a^x\ln a&\int a^x\,dx&=\frac{a^x}{\ln a}+C\end{aligned}\qquad(a>0,\ a\ne1)$$

5. 5. Basic trigonometric functions

   The familiar derivative identities supply the antiderivatives. In particular, differentiating cosine introduces a minus sign, so integrating sine needs one too:

   $$\frac{d}{dx}\sin x=\cos x,\qquad\frac{d}{dx}\cos x=-\sin x,\qquad\frac{d}{dx}\tan x=\sec^2x$$

   Reverse each identity to obtain the integration rules:

   $$\begin{aligned}\int\cos x\,dx&=\sin x+C\\[4pt]\int\sin x\,dx&=-\cos x+C\\[4pt]\int\sec^2x\,dx&=\tan x+C\end{aligned}$$

   These identities hold on intervals where the functions are defined; in particular, $\tan x$ and $\sec^2x$ are undefined where $\cos x=0$.

6. 6. Substitution reverses the chain rule

   If an inner function appears along with its derivative, treat the inner function as a new variable. The chain rule explains why this works:

   $$\begin{aligned}\frac{d}{dx}F(g(x))&=F'(g(x))g'(x)=f(g(x))g'(x)\\[4pt]\int f(g(x))g'(x)\,dx&=F(g(x))+C\end{aligned}$$

   For $u=x^2$, we have $du=2x\,dx$. The integral then becomes a familiar cosine rule:

   $$\int 2x\cos(x^2)\,dx=\int\cos u\,du=\sin u+C=\sin(x^2)+C$$

7. 7. Integration by parts reverses the product rule

   A product is different: differentiating it creates two terms. Integrate the product rule and move one term to the other side:

   $$\begin{aligned}\frac{d}{dx}(uv)&=u'v+uv'\\[4pt]d(uv)&=u\,dv+v\,du\\[4pt]\int u\,dv&=uv-\int v\,du\end{aligned}$$

   Choose $u=x$ and $dv=e^x\,dx$. Then $du=dx$ and $v=e^x$:

   $$\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x+C$$

8. 8. Why endpoint subtraction gives a definite integral

   Let $A(x)=\int_a^x f(t)\,dt$ be the area accumulated up to $x$. If $f$ is continuous, the extra area over a very short interval is approximately its width times $f(x)$. Shrinking that width gives:

   $$A'(x)=\lim_{h\to0}\frac{1}{h}\int_x^{x+h}f(t)\,dt=f(x)$$

   So $A$ and any antiderivative $F$ differ only by a constant. Because $A(a)=0$, that constant is $-F(a)$. At the upper bound:

   $$A(x)=F(x)-F(a)\qquad\Longrightarrow\qquad\int_a^b f(x)\,dx=F(b)-F(a)$$