Introduction to Integrals & Basic Rules
Antiderivative: A function F(x) is an antiderivative of f(x) if F'(x) = f(x).
Indefinite Integral: The set of all antiderivatives of f(x) is denoted by \int f(x) dx = F(x) + C, where F(x) is any antiderivative and C is the constant of integration.
\int dx = x + C
\int c\cdot f(x) dx = c \int f(x) dx
\int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx
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Why the + C?
Different functions can have the same derivative. For example, x^2, x^2+5, and x^2-10 all have a derivative of 2x. The constant C represents this arbitrary constant difference between any two antiderivatives.
Tip: Always remember to add the +C for indefinite integrals!
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Power Rule for Integration (n \ne -1)
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Integration of a Constant
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Example: \int \frac{2}{x} dx
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Integral of \csc(x)\cot(x)
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Techniques of Integration
Concept: Reverse of the Chain Rule. Used when you have an integrand of the form f(g(x))g'(x) dx.
Steps:
- Choose a substitution u = g(x).
- Compute the differential du = g'(x) dx.
- Rewrite the integral entirely in terms of u and du.
- Evaluate the integral with respect to u.
- Substitute back u = g(x) to express the result in terms of the original variable.
Example: \int x \cos(x^2) dx
- Let u = x^2.
- du = 2x dx, so x dx = \frac{1}{2} du.
- \int \cos(u) \frac{1}{2} du = \frac{1}{2} \int \cos(u) du.
- \frac{1}{2} \sin(u) + C.
- Substitute back: \frac{1}{2} \sin(x^2) + C.
Tip: Look for an inner function (g(x)) whose derivative (g'(x)) is also present in the integrand (up to a constant factor).
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Concept: Reverse of the Product Rule. Used for integrals of products of functions.
Formula: \int u \, dv = uv - \int v \, du
Mnemonic (LIATE): A guideline for choosing u in \int u \, dv. Choose u as the function type appearing first in this list:
- Logarithmic functions (\ln x, \log_b x)
- Inverse trigonometric functions (\arcsin x, \arctan x, etc.)
- Algebraic functions (x^n, polynomials)
- Trigonometric functions (\sin x, \cos x, etc.)
- Exponential functions (e^x, a^x)
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Example: \int x e^x dx
LIATE suggests u=x (Algebraic) and dv=e^x dx (Exponential).
- u = x \implies du = dx
- dv = e^x dx \implies v = \int e^x dx = e^x
Now apply the formula:
\int x e^x dx = x e^x - \int e^x dx
= x e^x - e^x + C
= e^x(x-1) + C
Example 2: \int \ln x dx
Let u = \ln x and dv = dx.
- u = \ln x \implies du = \frac{1}{x} dx
- dv = dx \implies v = x
Apply the formula:
\int \ln x dx = (\ln x)(x) - \int x \frac{1}{x} dx
= x \ln x - \int 1 dx
= x \ln x - x + C
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Involving Powers of \sin(x) and \cos(x): \int \sin^m(x) \cos^n(x) dx
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Case 1: n is odd. Save one \cos(x) factor and use \cos^2(x) = 1 - \sin^2(x) for the remaining factors. Let u = \sin(x), du = \cos(x) dx.
- Example: \int \sin^2(x) \cos^3(x) dx = \int \sin^2(x) (1-\sin^2(x)) \cos(x) dx. Let u = \sin(x), du = \cos(x) dx. \int u^2(1-u^2) du = \int (u^2-u^4) du = \frac{u^3}{3} - \frac{u^5}{5} + C = \frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} + C.
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Case 2: m is odd. Save one \sin(x) factor and use \sin^2(x) = 1 - \cos^2(x) for the remaining factors. Let u = \cos(x), du = -\sin(x) dx.
- Example: \int \sin^3(x) \cos^2(x) dx. Let u = \cos(x), du = -\sin(x) dx. \int (1-u^2)u^2 (-du) = \int (u^4-u^2) du = \frac{u^5}{5} - \frac{u^3}{3} + C = \frac{\cos^5(x)}{5} - \frac{\cos^3(x)}{3} + C.
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Case 3: Both m and n are even. Use half-angle identities: \sin^2(x) = \frac{1 - \cos(2x)}{2}, \cos^2(x) = \frac{1 + \cos(2x)}{2}.
- Example: \int \cos^2(x) dx = \int \frac{1 + \cos(2x)}{2} dx = \frac{1}{2} \int dx + \frac{1}{2} \int \cos(2x) dx = \frac{x}{2} + \frac{1}{4} \sin(2x) + C.
Involving Powers of \tan(x) and \sec(x): \int \tan^m(x) \sec^n(x) dx
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Case 1: n is even (n=2k, k \ge 1). Save a \sec^2(x) factor and use \sec^2(x) = 1 + \tan^2(x) for the remaining factors. Let u = \tan(x), du = \sec^2(x) dx.
- Example: \int \tan^3(x) \sec^4(x) dx = \int \tan^3(x) (1+\tan^2(x)) \sec^2(x) dx. Let u = \tan(x), du = \sec^2(x) dx. \int u^3(1+u^2) du = \int (u^3+u^5) du = \frac{u^4}{4} + \frac{u^6}{6} + C = \frac{\tan^4(x)}{4} + \frac{\tan^6(x)}{6} + C.
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Case 2: m is odd (m=2k+1). Save a \sec(x)\tan(x) factor and use \tan^2(x) = \sec^2(x) - 1 for the remaining factors. Let u = \sec(x), du = \sec(x)\tan(x) dx.
- Example: \int \tan^3(x) \sec(x) dx = \int (\sec^2(x)-1) \sec(x) \tan(x) dx. Let u = \sec(x), du = \sec(x)\tan(x) dx. \int (u^2-1) du = \frac{u^3}{3} - u + C = \frac{\sec^3(x)}{3} - \sec(x) + C.
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Other Cases: Often require reduction formulas or special tricks.
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Definite Integrals and Applications
FTC Part 1: If f is continuous on [a, b], then the function g(x) defined by g(x) = \int_a^x f(t) dt is continuous on [a, b] and differentiable on (a, b), and g'(x) = f(x).
\frac{d}{dx} \int_a^x f(t) dt = f(x)
Example: \frac{d}{dx} \int_1^x \sin(t^2) dt = \sin(x^2)
With Chain Rule: \frac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x)) \cdot g'(x)
Example: \frac{d}{dx} \int_0^{x^3} e^t dt = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2
FTC Part 2: If f is continuous on [a, b], then \int_a^b f(x) dx = F(b) - F(a), where F is any antiderivative of f, i.e., F'(x) = f(x).
\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)
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Steps:
- Find the indefinite integral (antiderivative) F(x) of the integrand f(x). Don’t need the +C here as it cancels out.
- Evaluate F(x) at the upper limit (b) and the lower limit (a).
- Subtract the value at the lower limit from the value at the upper limit: F(b) - F(a).
Example: \int_1^2 x^2 dx
- Antiderivative of x^2 is \frac{x^3}{3}. So F(x) = \frac{x^3}{3}.
- Evaluate at limits: F(2) = \frac{2^3}{3} = \frac{8}{3}, F(1) = \frac{1^3}{3} = \frac{1}{3}.
- Subtract: F(2) - F(1) = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.
Tip: If using U-substitution for a definite integral, you can change the limits of integration when you change the variable. If u = g(x), the new limits become g(a) and g(b). This avoids substituting back to x before evaluating.
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Concept: The area between two curves y=f(x) and y=g(x) from x=a to x=b, where f(x) \ge g(x) on [a, b].
Formula: Area = \int_a^b [f(x) - g(x)] dx
- f(x) is the upper curve, g(x) is the lower curve.
- If the upper/lower curve changes within the interval, you must split the integral into multiple integrals.
- Alternatively, integrate with respect to y: Area = \int_c^d [h(y) - k(y)] dy, where x=h(y) is the right curve and x=k(y) is the left curve, from y=c to y=d.
Example: Find the area bounded by y=x^2 and y=x for x \ge 0.
- Find intersection points: x^2 = x \implies x^2 - x = 0 \implies x(x-1) = 0 \implies x=0, x=1. So a=0, b=1.
- On [0, 1], y=x is the upper curve (f(x)=x) and y=x^2 is the lower curve (g(x)=x^2).
- Area = \int_0^1 [x - x^2] dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = (\frac{1^2}{2} - \frac{1^3}{3}) - (\frac{0^2}{2} - \frac{0^3}{3}) = (\frac{1}{2} - \frac{1}{3}) - 0 = \frac{3-2}{6} = \frac{1}{6}.
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Concept: Finding the volume of a solid of revolution formed by rotating a region about an axis.
Disk Method: Used when the region is adjacent to the axis of revolution, creating solid disks.
- Rotation about x-axis: V = \int_a^b \pi [f(x)]^2 dx
- Rotation about y-axis: V = \int_c^d \pi [g(y)]^2 dy (where x=g(y))
Washer Method: Used when there is a space between the region and the axis of revolution, creating washers (disks with holes).
- Rotation about x-axis: V = \int_a^b \pi ([R(x)]^2 - [r(x)]^2) dx, where R(x) is the outer radius and r(x) is the inner radius.
- Rotation about y-axis: V = \int_c^d \pi ([R(y)]^2 - [r(y)]^2) dy, where R(y) is the outer radius and r(y) is the inner radius.
Example (Disk): Volume of solid formed by rotating the region under y=\sqrt{x} from x=0 to x=4 about the x-axis.
- a=0, b=4, f(x) = \sqrt{x}.
- V = \int_0^4 \pi (\sqrt{x})^2 dx = \pi \int_0^4 x dx = \pi [\frac{x^2}{2}]_0^4 = \pi (\frac{4^2}{2} - \frac{0^2}{2}) = \pi (8 - 0) = 8\pi.
Example (Washer): Volume of solid formed by rotating the region bounded by y=x^2 and y=x (x \ge 0) about the x-axis.
- Intersection points at x=0, x=1. So a=0, b=1.
- Outer radius R(x) is the function farther from the axis, y=x. So R(x) = x.
- Inner radius r(x) is the function closer to the axis, y=x^2. So r(x) = x^2.
- V = \int_0^1 \pi [x^2 - (x^2)^2] dx = \pi \int_0^1 [x^2 - x^4] dx = \pi [\frac{x^3}{3} - \frac{x^5}{5}]_0^1 = \pi ((\frac{1^3}{3} - \frac{1^5}{5}) - (0-0)) = \pi (\frac{1}{3} - \frac{1}{5}) = \pi (\frac{5-3}{15}) = \frac{2\pi}{15}.
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