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Reversing the power rule and the derivative table, substitution as the chain rule read backwards, and the areas and volumes that a definite integral measures.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can find antiderivatives by reversing the power rule and the derivative table, and use substitution — choosing $u$ so that its derivative is already present, and changing the limits with it. You can also compute the area between two curves, the volume of a solid of revolution by discs or washers, and the volume of a solid with known cross-sections, all by describing one thin slice and integrating it.
You know the fundamental theorem, so you know that evaluating an integral means finding an antiderivative. This section is about how to find one — and about the single technique, substitution, that turns the chain rule around.
Antiderivative: a function whose derivative is the integrand.
Indefinite integral $\int f\,dx$: all of them, written $F(x) + C$.
Substitution: replacing the inside of a composite by $u$, so the integral becomes one in $u$.
Changing the limits: converting $x$ limits to $u$ limits, so no substituting back is needed.
Reversing the power rule. Since differentiating multiplies by the exponent and drops it, antidifferentiating raises it and divides:
$$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \ne -1)$$
The exception matters: $n = -1$ would divide by zero, and $\int \frac{1}{x}dx = \ln|x| + C$ instead — which is where logarithms enter integration.
The rest of the table, backwards. $\int e^x dx = e^x + C$; $\int \cos x\,dx = \sin x + C$; and $\int \sin x\,dx = -\cos x + C$. That last minus sign is the one that moved: it belonged to the derivative of $\cos$, and reversing the table attaches it to the antiderivative of $\sin$.
The $+C$. Differentiation destroys constants, so antidifferentiation cannot recover one. An indefinite integral names a whole family of functions, all parallel, and $C$ says which. In a definite integral the constant appears at both limits and cancels, which is why it is dropped there — and only there.
Substitution. This is the chain rule read backwards. Choose $u$ to be the inside of a composite; then $du = u'(x)\,dx$, and the integral turns into one in $u$ alone. The method works exactly when $u'$ is already present as a factor, up to a constant — so the check is not optional, it is the method. For a definite integral, change the limits to $u$ values at the same time, and there is nothing to substitute back.
Another way: picture
Picture the derivative table read from right to left. Every entry that was an answer is now a question. $\cos x$ came from $\sin x$; $e^x$ came from $e^x$; $\tfrac{1}{x}$ came from $\ln x$. Antidifferentiation is recognition rather than calculation, which is why it is harder.
Another way: steps
Dropping the $+C$. An indefinite integral is a family of functions, not one function. In a differential equation the missing constant is precisely what the initial condition is there to determine.
Substituting without checking $du$. $u$ must be chosen so that its derivative already appears as a factor, up to a constant. If it does not, an $x$ survives the substitution and the integral cannot be finished.
Keeping the old limits. After substituting, $\int_0^2$ means "from $u = 0$ to $u = 2$", which is almost never what was meant. Either convert the limits or substitute back before evaluating — never neither.
$\int 2x(x^2+1)^3 dx$: let $u = x^2 + 1$, so $du = 2x\,dx$.
The $2x$ is already there.
The integral becomes $\int u^3 du = \dfrac{u^4}{4} + C$.
An ordinary power now.
Substitute back: $\dfrac{(x^2+1)^4}{4} + C$.
$\int_0^1 2x(x^2+1)^3 dx$, $u = x^2+1$: $x = 0 \to u = 1$, $x = 1 \to u = 2$.
Convert the limits.
$\int_1^2 u^3 du = \left[\dfrac{u^4}{4}\right]_1^2 = \dfrac{16 - 1}{4} = \dfrac{15}{4}$.
No substituting back was needed — and no chance of using $u$'s integrand with $x$'s limits.
$u = x^3 + 5$, so $du = 3x^2 dx$ — present already.
$\int u^2 du = \dfrac{u^3}{3} + C$.
$\dfrac{(x^3+5)^3}{3} + C$, which differentiates back to the original.
Find $\displaystyle\int (9x^2 + 10x) \, dx$. Write the constant of integration as $C$.
Answer:
For $\displaystyle\int 2x(3x^2 + 3)^4 \, dx$, what should $u$ be?
Evaluate $\displaystyle\int_0^{2} 2x(x^2 + 3)^2 \, dx$.
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$\displaystyle\int_0^{\pi/2} 7\cos x \, dx$. What is the value?
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You can evaluate a definite integral, and you know it measures a signed area. This section uses that to measure regions between curves and solids built from them — and every case reduces to the same question: what does one thin slice look like?
Area between curves: $\int_a^b (\text{top} - \text{bottom})\,dx$.
Solid of revolution: what a region sweeps out when rotated about a line.
Disc: a circular slice, area $\pi r^2$.
Washer: a disc with a concentric hole, area $\pi(R^2 - r^2)$.
Known cross-sections: slices of a stated shape, integrated along an axis.
Every problem in this section is the same problem: describe one thin slice, then integrate its measure along the appropriate axis.
Area between two curves. A vertical strip at $x$ has height $(\text{top} - \text{bottom})$ and thickness $dx$, so the area is $\int_a^b (\text{top} - \text{bottom})\,dx$ across the interval where they overlap. If the curves cross, split the integral at the crossing — the words 'top' and 'bottom' change places there.
Volume by discs. Rotate a region against the axis and each slice perpendicular to the axis is a disc of radius $f(x)$, so the volume is $\int_a^b \pi f(x)^2\,dx$. The square is inside the integral, always.
Volume by washers. If the region does not touch the axis, the slice has a hole: $\int_a^b \pi(R(x)^2 - r(x)^2)\,dx$, with $R$ the outer radius and $r$ the inner. Square each, then subtract.
Known cross-sections. If the slices are squares, triangles or semicircles rather than discs, integrate whatever their area is. Discs and washers are the special case of this rule, not a separate one — and no $\pi$ appears unless a circle does.
Sometimes the natural strip is horizontal, with thickness $dy$ and length measured in $x$. Choosing the direction of the strip before writing anything is what keeps a problem from needing to be split into pieces.
Another way: picture
Picture a loaf of bread. Its volume is the area of one slice times the thickness, added over all the slices — and if the slices are different sizes, that sum is an integral. Discs, washers and squares are just different loaves.
Another way: steps
| Problem | One slice | Integrate |
|---|---|---|
| Area between curves | A strip, height top $-$ bottom | $\int (\text{top} - \text{bottom})\,dx$ |
| Rotated, region touches the axis | A disc, radius $f(x)$ | $\int \pi f^2\,dx$ |
| Rotated, region away from the axis | A washer | $\int \pi(R^2 - r^2)\,dx$ |
| Known cross-sections | That shape | $\int A(x)\,dx$ |
The fourth row contains the other three. Naming the slice is the whole of the method; the integration afterwards is routine.
Bottom minus top. The order decides the sign. A negative area is not a negative area — it is the right number with the subtraction the wrong way round.
$(R - r)^2$ instead of $R^2 - r^2$. A washer is the difference of two areas, so each radius is squared before subtracting. These are not equal, and the mistake is invisible in the final number.
A $\pi$ where nothing was rotated. A solid with square cross-sections has no circle anywhere in it. $\pi$ comes from the shape of a slice, not from the fact that a volume is being computed.
$y = 4x$ and $y = x^2$ meet where $x^2 = 4x$, at $x = 0$ and $x = 4$.
Find the limits first.
Between them the line is above, so integrate $4x - x^2$ from $0$ to $4$.
Top minus bottom.
$\left[2x^2 - \dfrac{x^3}{3}\right]_0^4 = 32 - \dfrac{64}{3} = \dfrac{32}{3}$.
Rotate the region under $y = 2x$ from $0$ to $3$ about the $x$-axis.
A cone of radius 6, height 3.
$\int_0^3 \pi(2x)^2 dx = \pi\int_0^3 4x^2 dx = 36\pi$.
Square inside the integral.
The formula $\tfrac{1}{3}\pi r^2 h$ gives $\tfrac{1}{3}\pi(36)(3) = 36\pi$ too — and this integral is where that formula comes from.
They meet at $0$ and $1$, and $x \ge x^2$ between them.
Sketch first.
$\int_0^1 (x - x^2)dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1$.
$= \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}$ — small, as the picture suggests it should be.
Find the area between $y = 3x$ and $y = x^2$ from $x = 0$ to $x = 3$, where the line is above.
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The region under $y = 3x$ from $x = 0$ to $x = 1$ is rotated about the $x$-axis. Find the volume, as a multiple of $\pi$.
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The region between $y = 3$ and $y = 1$, from $x = 0$ to $x = 3$, is rotated about the $x$-axis. Find the volume, as a multiple of $\pi$.
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Find the area enclosed by $x = 3y$, the $y$-axis, and the line $y = 2$, by integrating with respect to $y$.
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Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Why does $\displaystyle\int 12x \, dx = 6x^2 + C$ carry a $+ C$, when $\displaystyle\int_0^1 12x \, dx = 6$ does not?
A solid has a base in the $xy$-plane, and every cross-section perpendicular to the $x$-axis is a square of side $4x$. What integral gives its volume from $0$ to $1$?
You can antidifferentiate and compute areas and volumes. Without looking: what has to be true of $u$ before a substitution will work, and why is a washer $\pi(R^2 - r^2)$ rather than $\pi(R - r)^2$?
8. Your turn: $\int 3x^2(x^3 + 5)^2 dx$, step 3
20. Your turn: the region between $y = x$ and $y = x^2$ from $0$ to $1$, step 3