Integrals: the basics

Signed area

A definite integral is a signed area: it is negative on intervals where the graph lies below the x-axis.

f(x)≤0 on [a,b]⇒∫abf(x) dx≤0\begin{gathered} f(x)\le 0 \text{ on } [a,b] \\[4pt] \Rightarrow \int_a^b f(x)\,dx\le 0 \end{gathered}

A definite integral is a signed area. On intervals where the graph is below the xx-axis, the value is negative.

To find the actual area, split the interval at the points where the graph crosses the xx-axis, flip the sign of the negative parts, and add.

xyy = x² − 102
The part below the x-axis (blue) counts as negative. For the actual area, split at the crossing and flip the sign.Drag the dot

Examples

  1. 01What is the value of this definite integral?∫−22x3 dx\displaystyle \int_{-2}^{2} x^{3}\,dxAnswerHide
    Answer0\displaystyle 0

    Steps

    1. ∫−20x3 dx\displaystyle \int_{-2}^{0} x^{3}\,dx=−4\displaystyle = -4∫02x3 dx\displaystyle \int_{0}^{2} x^{3}\,dx=4\displaystyle = 4
    2. Adding them cancels
    3. ∫−22x3 dx\displaystyle \int_{-2}^{2} x^{3}\,dx=0\displaystyle = 0
  2. 02What is the value of this definite integral?∫−33x3 dx\displaystyle \int_{-3}^{3} x^{3}\,dxAnswerHide
    Answer0\displaystyle 0

    Steps

    1. ∫−30x3 dx\displaystyle \int_{-3}^{0} x^{3}\,dx=−814\displaystyle = -\frac{81}{4}∫03x3 dx\displaystyle \int_{0}^{3} x^{3}\,dx=814\displaystyle = \frac{81}{4}
    2. Adding them cancels
    3. ∫−33x3 dx\displaystyle \int_{-3}^{3} x^{3}\,dx=0\displaystyle = 0
  3. 03What is the value of this definite integral?∫−11x3 dx\displaystyle \int_{-1}^{1} x^{3}\,dxAnswerHide
    Answer0\displaystyle 0

    Steps

    1. ∫−10x3 dx\displaystyle \int_{-1}^{0} x^{3}\,dx=−14\displaystyle = -\frac{1}{4}∫01x3 dx\displaystyle \int_{0}^{1} x^{3}\,dx=14\displaystyle = \frac{1}{4}
    2. Adding them cancels
    3. ∫−11x3 dx\displaystyle \int_{-1}^{1} x^{3}\,dx=0\displaystyle = 0
  4. 04What is the value of this definite integral?∫−11x1 dx\displaystyle \int_{-1}^{1} x^{1}\,dxAnswerHide
    Answer0\displaystyle 0

    Steps

    1. ∫−10x1 dx\displaystyle \int_{-1}^{0} x^{1}\,dx=−12\displaystyle = -\frac{1}{2}∫01x1 dx\displaystyle \int_{0}^{1} x^{1}\,dx=12\displaystyle = \frac{1}{2}
    2. Adding them cancels
    3. ∫−11x1 dx\displaystyle \int_{-1}^{1} x^{1}\,dx=0\displaystyle = 0
  5. 05What is the value of this definite integral?∫−33x1 dx\displaystyle \int_{-3}^{3} x^{1}\,dxAnswerHide
    Answer0\displaystyle 0

    Steps

    1. ∫−30x1 dx\displaystyle \int_{-3}^{0} x^{1}\,dx=−92\displaystyle = -\frac{9}{2}∫03x1 dx\displaystyle \int_{0}^{3} x^{1}\,dx=92\displaystyle = \frac{9}{2}
    2. Adding them cancels
    3. ∫−33x1 dx\displaystyle \int_{-3}^{3} x^{1}\,dx=0\displaystyle = 0
  6. 06What is the value of this definite integral?∫−22x1 dx\displaystyle \int_{-2}^{2} x^{1}\,dxAnswerHide
    Answer0\displaystyle 0

    Steps

    1. ∫−20x1 dx\displaystyle \int_{-2}^{0} x^{1}\,dx=−2\displaystyle = -2∫02x1 dx\displaystyle \int_{0}^{2} x^{1}\,dx=2\displaystyle = 2
    2. Adding them cancels
    3. ∫−22x1 dx\displaystyle \int_{-2}^{2} x^{1}\,dx=0\displaystyle = 0

FAQ

When is a definite integral negative?
When the graph is below the x-axis on that interval (f(x)≤0f(x)\le 0). If you are asked for the area, fix the sign of that part and add it.

Published 2026-09-03 · Updated 2026-09-05