A structured guide to evaluating and approximating integrals using antiderivatives, the Fundamental Theorem of Calculus, substitution, algebraic and trigonometric techniques, and numerical rules.
Antiderivatives and basic formulas
Integration measures accumulated change and reverses differentiation when an exists. It can describe signed area, displacement, and other accumulated quantities.
An of f is a function F satisfying
F′(x)=f(x).
Because constants disappear under differentiation, all antiderivatives of f have the form F(x)+C. This gives the :
∫f(x)dx=F(x)+C.
Important basic formulas include
∫xndx=n+1xn+1+C(n=−1),
∫x1dx=ln∣x∣+C,∫exdx=ex+C,
∫cosxdx=sinx+C,∫sinxdx=−cosx+C.
The constant-multiple and sum rules allow an integrand to be separated:
∫kf(x)dx=k∫f(x)dx,
∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx.
For example,
∫(3x2−4x+5)dx=x3−2x2+5x+C.
Differentiating the result is a direct check.
Takeaway: Find an , include C for an , and differentiate the result to verify it.
Definite integrals and signed accumulation
A has limits of integration and produces a number:
∫abf(x)dx.
It represents signed accumulation from x=a to x=b. Regions where the function is below the x-axis contribute negatively, so signed accumulation can differ from total geometric area.
Useful properties are
∫aaf(x)dx=0,
∫abf(x)dx=−∫baf(x)dx,
∫acf(x)dx+∫cbf(x)dx=∫abf(x)dx,
and
∫ab[kf(x)+g(x)]dx=k∫abf(x)dx+∫abg(x)dx.
A does not include C, because any constant cancels when the endpoint values are subtracted.
Takeaway: Pay attention to orientation, signs, and whether the problem asks for signed accumulation or geometric area.
The Fundamental Theorem in action
The has two complementary parts.
First, an accumulation function has the original integrand as its derivative. If f is continuous and
F(x)=∫axf(t)dt,
then
F′(x)=f(x).
The variable t is a dummy variable that keeps the variable endpoint x separate from the variable of integration. If the upper endpoint is itself a function, the chain rule is also required:
G(x)=∫au(x)f(t)dt⟹G′(x)=f(u(x))u′(x).
For example,
dxd(∫1x21+t3dt)=2x1+x6.
Second, a can be evaluated using any F:
∫abf(x)dx=F(b)−F(a)=F(x)∣ab.
For instance,
∫02(x2+1)dx=[3x3+x]02=314.
Takeaway: Differentiation undoes accumulation, while endpoint subtraction evaluates accumulated change.
Reversing the chain and product rules
is most effective when an integrand contains a composite expression together with its derivative. Set
u=g(x),du=g′(x)dx.
Then rewrite the entire integral in terms of u, integrate, and substitute back if the integral is indefinite.
For example,
∫6x(3x2+4)4dx
uses u=3x2+4 and du=6xdx. Therefore,
∫6x(3x2+4)4dx=∫u4du=5u5+C=5(3x2+4)5+C.
For a , either return to x before applying the original limits or transform the limits:
∫abf(g(x))g′(x)dx=∫g(a)g(b)f(u)du.
is based on the product rule:
∫udv=uv−∫vdu.
Choose u so that differentiating it simplifies the expression, and choose dv so that it can be integrated readily. A common guideline is LIATE: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential.
For example, with u=x and dv=cosxdx, we have du=dx and v=sinx. Thus,
∫xcosxdx=xsinx−∫sinxdx=xsinx+cosx+C.
Takeaway: Use for a chain-rule pattern and for products or factors that simplify when differentiated.
Trigonometric integrals and identities
Trigonometric identities often turn a difficult trigonometric integral into a or a standard formula.
For
∫sinmxcosnxdx,
save one factor of sinx when m is odd and convert the remaining even power using sin2x=1−cos2x. When n is odd, save one factor of cosx and use cos2x=1−sin2x. If both powers are even, use
sin2x=21−cos2x,cos2x=21+cos2x.
For example,
∫sin3xcos2xdx
can be rewritten as sinx(1−cos2x)cos2x. With u=cosx, the result is
−3cos3x+5cos5x+C.
For tangent and secant, use
1+tan2x=sec2x.
An even power of secx suggests reserving sec2x for u=tanx, while an odd power of tanx can suggest reserving secxtanx for u=secx. Frequently used results include
∫tanxdx=−ln∣cosx∣+C,
∫secxdx=ln∣secx+tanx∣+C.
Takeaway: Inspect parity first, preserve a factor that matches a derivative, and use identities to convert the remaining powers.
Radicals and rational functions
is designed for radicals with one of three quadratic patterns:
For a2−x2, use x=asinθ and 1−sin2θ=cos2θ.
For a2+x2, use x=atanθ and 1+tan2θ=sec2θ.
For x2−a2, use x=asecθ and sec2θ−1=tan2θ.
After integrating in θ, use a right triangle or the original to return to x.
For example, consider
∫9+x2dx.
Set x=3tanθ. Then dx=3sec2θdθ and 9+x2=3secθ, so
∫9+x2dx=∫secθdθ=ln∣secθ+tanθ∣+C.
Since tanθ=x/3 and secθ=x2+9/3, the result is
∫9+x2dx=lnx+x2+9+C.
applies to rational functions after polynomial division when the numerator degree is not less than the denominator degree. Decompose the rational function according to its factor types, solve for the unknown coefficients, and integrate each simpler term.
For example,
(x−1)(x+2)5x+1=x−12+x+23,
so
∫(x−1)(x+2)5x+1dx=2ln∣x−1∣+3ln∣x+2∣+C.
Takeaway: Match each radical or denominator structure to its standard decomposition or pattern.
Numerical methods and technique selection
estimates a from function values. Divide [a,b] into n subintervals using
Δx=nb−a,xi=a+iΔx.
The midpoint rule uses the center of each subinterval:
Mn=Δxi=1∑nf(2xi−1+xi).
The trapezoidal rule connects neighboring points with line segments:
For ∫01x2dx with the trapezoidal rule and n=4, Δx=1/4, and the estimate is
T4=3211=0.34375.
The exact value is
∫01x2dx=31≈0.33333,
so the absolute error is approximately 0.01042. Increasing n generally improves the estimate, and Simpson’s rule is often more accurate for smooth functions.
When selecting a technique, simplify first, then look for a direct formula, , , trigonometric identities, , , or . Always check an by differentiating it and check definite-integral signs and endpoint values.
Takeaway: Use exact techniques when practical; use numerical rules when an exact is unavailable or unnecessary, and compare estimates with exact values when possible.