Calculus

Integration for Beginners: Antiderivatives and Area

By Math Solving Space · · 2 min read

On this page
  1. Integration undoes differentiation
  2. The power rule for integrals
  3. Definite integrals and area
  4. Standard integrals
  5. Integration by parts
  6. Always check by differentiating
  7. Common mistakes
  8. Practice questions
  9. Frequently asked questions

Integration is the reverse of differentiation. An indefinite integral finds a function whose derivative is the one you started with, plus a constant CC. A definite integral between two limits gives the signed area under the curve. Most beginner integrals use the power rule and a short table of standard results.

Integration undoes differentiation

Since ddxx3=3x2\frac{d}{dx}x^3 = 3x^2, it follows that

∫3x2 dx=x3+C\int 3x^2\,dx = x^3 + C

The +C+ C is there because x3+5x^3 + 5 and x3−2x^3 - 2 also have derivative 3x23x^2.

The power rule for integrals

∫xn dx=xn+1n+1+C(n≠−1)\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \ne -1)

Raise the power by one, then divide by the new power. The exception is n=−1n = -1:

∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C

Definite integrals and area

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)

where FF is any antiderivative. Example:

∫03x2 dx=[x33]03=273−0=9\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9

So the area under y=x2y = x^2 from 00 to 33 is exactly 99 square units.

Another example: ∫0πsin⁡x dx=[−cos⁡x]0π=1−(−1)=2\int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = 1 - (-1) = 2.

Standard integrals

f(x)f(x) ∫f(x) dx\int f(x)\,dx
xnx^n xn+1n+1\frac{x^{n+1}}{n+1}
exe^x exe^x
sin⁡x\sin x −cos⁡x-\cos x
cos⁡x\cos x sin⁡x\sin x
11+x2\frac{1}{1+x^2} arctan⁡x\arctan x

For a linear inside such as cos⁡(3x)\cos(3x), divide by the coefficient: ∫cos⁡(3x) dx=13sin⁡(3x)+C\int\cos(3x)\,dx = \frac{1}{3}\sin(3x) + C.

Integration by parts

For a product such as xexx e^x:

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du

Choose u=xu = x (it gets simpler when differentiated) and dv=ex dxdv = e^x\,dx. Then du=dxdu = dx and v=exv = e^x:

∫xex dx=xex−∫ex dx=(x−1)ex+C\int x e^x\,dx = x e^x - \int e^x\,dx = (x - 1)e^x + C

Always check by differentiating

Differentiate your answer: ddx(x−1)ex=ex+(x−1)ex=xex\frac{d}{dx}(x-1)e^x = e^x + (x-1)e^x = xe^x ✓. This one habit catches most integration mistakes.

Common mistakes

  • Forgetting +C+ C on indefinite integrals.
  • Dividing by the old power instead of the new one.
  • Getting the sign of ∫sin⁡x dx\int \sin x\,dx wrong (it is −cos⁡x-\cos x).

Practice questions

  1. ∫(4x3+2x) dx\int (4x^3 + 2x)\,dx
  2. ∫12x dx\int_1^2 x\,dx
  3. ∫e2x dx\int e^{2x}\,dx

Answers: 1) x4+x2+Cx^4 + x^2 + C 2) 32\frac{3}{2} 3) 12e2x+C\frac{1}{2}e^{2x} + C

The integral calculator shows each rule, shades the area for definite integrals and animates the rectangles that approach it. Revise the derivative rules first if they are rusty.

Frequently asked questions

Why do we add + C?

Because the derivative of any constant is zero, so infinitely many functions share the same derivative.

What is the difference between definite and indefinite integrals?

An indefinite integral is a family of functions. A definite integral has limits and gives a number, the signed area.

Can every function be integrated?

Every continuous function has an antiderivative, but some, like e^(−x²), cannot be written with ordinary functions. Their definite integrals are found numerically.

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