The Fundamental Theorem of Calculus and substitution

The most important theorem of Calculus is this week’s. The areas you computed at length with Riemann sums come down to one line, because integration and differentiation are inverses of each other. The theorem has two parts, and on exams both appear as separate question types. The second half of the week is substitution: the integral counterpart of the chain rule for derivatives. In every integral question from now on, this will be the first method you try.

1. $\dfrac{d}{dx}\displaystyle\int_a^xf(t)\,dt=f(x)$. If the upper limit is $g(x)$, multiply by $g'(x)$.
2. $\displaystyle\int_a^bf\,dx=F(b)-F(a)$. First the antiderivative, then upper minus lower.
3. In substitution, choose $u$ as the inner expression whose derivative is also in the integral.
4. In a definite integral, convert the limits to $u$ as well.

The Fundamental Theorem of Calculus, Part 1

Let $f$ be continuous on $[a,b]$ and define

$g(x)=\int_a^xf(t)\,dt$

$g(x)$ is the (signed) area accumulated from $a$ to $x$; it is called the accumulation function. The theorem:

$g'(x)=f(x)$

Intuition: if you increase $x$ a little, by $h$, a thin strip of width $h$ and height about $f(x)$ is added to the area. The area increases by $f(x)\cdot h$, and the rate of increase is $f(x)$.

The general form with the chain rule:

$\frac{d}{dx}\int_a^{g(x)}f(t)\,dt=f\big(g(x)\big)\,g'(x)$

Example 1. $\dfrac{d}{dx}\displaystyle\int_1^x\cos(t^2)\,dt=\cos(x^2)$. The antiderivative of $\cos(t^2)$ cannot be written with elementary functions, but it is not needed to answer a derivative question.

Example 2. $\dfrac{d}{dx}\displaystyle\int_1^{x^2}\ln t\,dt=\ln(x^2)\cdot2x=4x\ln x$ ($x>0$).

If the lower limit is the variable, flip the sign: $\displaystyle\int_{h(x)}^{a}f=-\int_a^{h(x)}f$. If both limits are variable, put a fixed point in the middle and split in two.

PITFALL · forgetting the chain factor. Writing $\dfrac{d}{dx}\int_0^{x^3}\sin t\,dt=\sin(x^3)$ is wrong. The correct answer is $3x^2\sin(x^3)$. If the upper limit is not $x$, multiply by its derivative.

The Fundamental Theorem of Calculus, Part 2

If $F$ is any antiderivative of $f$:

$\int_a^bf(x)\,dx=F(b)-F(a)=\Big[F(x)\Big]_a^b$

You don’t need to write $+C$: it cancels in the difference $(F(b)+C)-(F(a)+C)$.

Example 3. $\displaystyle\int_1^2\left(x^2-\frac1x\right)dx=\left[\frac{x^3}{3}-\ln x\right]_1^2=\left(\frac83-\ln2\right)-\left(\frac13-0\right)=\frac73-\ln2\approx1.640$.

Example 4. $\displaystyle\int_0^\pi\sin x\,dx=\big[-\cos x\big]_0^\pi=-\cos\pi+\cos0=1+1=2$. The area under one arch of the sine is exactly $2$.

Example 5 (absolute value). $\displaystyle\int_{-1}^2|x|\,dx=\int_{-1}^0(-x)\,dx+\int_0^2x\,dx=\frac12+2=\frac52$. You cannot find an antiderivative of the absolute value without splitting it at the zero point.

Why are the two parts the same theorem?

Part 2 follows from Part 1. The derivative of the function $g(x)=\int_a^xf(t)\,dt$ is $f$, so $g$ is also an antiderivative of $f$. Since every antiderivative differs by a constant (week 9, MVT consequence), $F(x)=g(x)+C$. Now $F(b)-F(a)=g(b)-g(a)=\int_a^bf-0$. The area defined by the limit of Riemann sums can be computed with the inverse of the derivative. This is the sentence that turns Calculus into a tool.

In practice it means: in week 11, for $\int_0^1x^2\,dx$ you wrote half a page of summation formulas. Now it is $\left[\tfrac{x^3}{3}\right]_0^1=\tfrac13$, one line.

Net change and average value

$\displaystyle\int_a^bF'(x)\,dx=F(b)-F(a)$: the integral of a rate of change gives the total (net) change. The integral of velocity is displacement, and the integral of flow rate is total volume.

Average value:

$f_{\text{avg}}=\frac{1}{b-a}\int_a^bf(x)\,dx$

Mean Value Theorem for integrals: if $f$ is continuous, there is a $c\in[a,b]$ with $f(c)=f_{\text{avg}}$. For example the average of $\sin x$ on $[0,\pi]$ is $\tfrac2\pi\approx0.637$.

Substitution

The chain rule said $\dfrac{d}{dx}F(g(x))=F'(g(x))g'(x)$. Read backwards:

$\int f\big(g(x)\big)\,g'(x)\,dx=\int f(u)\,du,\qquad u=g(x),\ du=g'(x)\,dx$

When it works: if the integrand contains an inner function and its derivative (a constant factor doesn’t matter) together.

Steps

1. Choose u: usually the inside of parentheses, inside a root, an exponent, a denominator
2. Compute du = u' dx, write dx or "the part with x" in terms of du
3. Write the integral entirely in terms of u (no x should remain)
4. Integrate with respect to u
5. Replace u with g(x) (for an indefinite integral)

Example 6. $\displaystyle\int2x\cos(x^2)\,dx$. $u=x^2$, $du=2x\,dx$. $\int\cos u\,du=\sin u+C=\sin(x^2)+C$.

Example 7 (linear inner function). $\displaystyle\int(3x+1)^5\,dx$. $u=3x+1$, $du=3\,dx$, $dx=\tfrac13du$. $\tfrac13\cdot\tfrac{u^6}{6}=\dfrac{(3x+1)^6}{18}+C$. Short rule: $\int f(ax+b)\,dx=\tfrac1aF(ax+b)+C$.

Example 8 ($\tan x$). $\displaystyle\int\tan x\,dx=\int\frac{\sin x}{\cos x}dx$. $u=\cos x$, $du=-\sin x\,dx$. $-\int\dfrac{du}{u}=-\ln|u|+C=-\ln|\cos x|+C=\ln|\sec x|+C$.

Example 9. $\displaystyle\int\frac{e^x}{1+e^{2x}}dx$. $u=e^x$, $du=e^xdx$, $e^{2x}=u^2$. $\displaystyle\int\frac{du}{1+u^2}=\arctan u+C=\arctan(e^x)+C$.

Which $u$?

Example 9b (back-substitution). $\displaystyle\int x\sqrt{x+1}\,dx$. The $x$ outside is not the derivative of $x+1$. Still choose $u=x+1$; $du=dx$ and $x=u-1$:

$\int(u-1)\sqrt u\,du=\int\left(u^{3/2}-u^{1/2}\right)du=\frac25u^{5/2}-\frac23u^{3/2}+C=\frac25(x+1)^{5/2}-\frac23(x+1)^{3/2}+C$

If you can write the remaining $x$ in terms of $u$, the method still works. Question 10 is also of this type.

PITFALL · leaving $x$ in the integral. For $\int x^2\cos(x)\,dx$, if you choose $u=x^2$ then $du=2x\,dx$, but $\cos x$ and an extra $x$ remain in the integrand, and it cannot be written in terms of $u$. Substitution does not solve every integral; this integral is solved with integration by parts in week 13.

Substitution in a definite integral

Convert the limits to $u$ as well. That way you don’t need to return to $x$ at the end.

Example 10. $\displaystyle\int_0^1x\sqrt{1-x^2}\,dx$.

u = 1 - x^2,   du = -2x dx   ->   x dx = -du/2
x = 0 -> u = 1,    x = 1 -> u = 0
∫_1^0 sqrt(u) (-du/2) = (1/2) ∫_0^1 u^(1/2) du = (1/2)(2/3) = 1/3

When the limits were swapped, the minus sign disappeared. This is the property $\int_b^a=-\int_a^b$.

Symmetry

If $f$ is odd, $\displaystyle\int_{-a}^af=0$; if it is even, $\displaystyle\int_{-a}^af=2\int_0^af$. For example $\displaystyle\int_{-2}^2x^3\cos x\,dx=0$, because $x^3$ is odd, $\cos x$ is even, and their product is odd. The answer without any computation.

Common mistakes