Integrals: functions and by parts

Integration by parts for definite integrals

Integration by parts for definite integrals is ∫abuv′ dx=[uv]ab−∫abu′v dx\int_{a}^{b} uv'\,dx=\left[uv\right]_{a}^{b}-\int_{a}^{b} u'v\,dx; the uvuv term is evaluated at the limits as well.

∫abuv′ dx=[uv]ab−∫abu′v dx\begin{gathered} \int_{a}^{b} uv'\,dx \\[4pt] =\left[uv\right]_{a}^{b}-\int_{a}^{b} u'v\,dx \end{gathered}

As with indefinite integrals, choose uu as the factor that gets simpler when differentiated and v′v' as the one that is easy to integrate. The difference is that the uvuv term is also evaluated at aa and bb and becomes a number.

∫01xex dx=[xex]01−∫01ex dx=e−(e−1)=1\begin{gathered} \int_{0}^{1} xe^{x}\,dx \\[4pt] =\left[xe^{x}\right]_{0}^{1}-\int_{0}^{1} e^{x}\,dx \\[4pt] =e-(e-1) \\[4pt] =1 \end{gathered}

Here u=xu=x and v′=exv'=e^{x}, so v=exv=e^{x}. Compute [xex]01=e\left[xe^{x}\right]_{0}^{1}=e and ∫01ex dx=e−1\int_{0}^{1}e^{x}\,dx=e-1 separately, then subtract.

Examples

  1. 01What is the value?∫01xex dx=[xex]01−∫01ex dx\displaystyle \int_{0}^{1} xe^{x}\,dx = [xe^{x}]_{0}^{1} - \int_{0}^{1} e^{x}\,dxAnswerHide
    Answer1\displaystyle 1

    Steps

    1. The [uv][uv] part takes its limits first and becomes a number
    2. 1⋅e1−0⋅e0\displaystyle 1 \cdot e^{1} - 0 \cdot e^{0}=e\displaystyle = e
    3. The remaining integral becomes a number too
    4. ∫01ex dx\displaystyle \int_{0}^{1} e^{x}\,dx=e−1\displaystyle = e-1
    5. Subtract; forgetting to do so leaves ee
    6. e−(e−1)\displaystyle e-(e-1)=1\displaystyle = 1
  2. 02What is v?u=x,v′=cos⁡x\displaystyle u=x, \quad v'=\cos xAnswerHide
    Answersin⁡x\displaystyle \sin x

    Steps

    1. vv is the integral of v′v', and integrating cos⁡x\cos x keeps the sign
    2. ∫cos⁡x dx\displaystyle \int \cos x\,dx=sin⁡x+C\displaystyle = \sin x+C
    3. The minus appears when you integrate sin⁡\sin, not cos⁡\cos
  3. 03What goes in □?∫abuv′ dx=[uv]ab−∫ab□ dx\displaystyle \int_{a}^{b} uv'\,dx = [uv]_{a}^{b} - \int_{a}^{b} \square\,dxAnswerHide
    Answeru′v\displaystyle u'v

    Steps

    1. The formula is the product rule (uv)′=u′v+uv′(uv)'=u'v+uv', rearranged for uv′uv'
    2. Integrating uv′=(uv)′−u′vuv'=(uv)'-u'v turns ∫(uv)′\int (uv)' into [uv][uv] and leaves −∫u′v-\int u'v
    3. The prime moves across: what you subtract is u′vu'v, not uv′uv' again
  4. 04What is the value?∫01ex dx\displaystyle \int_{0}^{1} e^{x}\,dxAnswerHide
    Answere−1\displaystyle e-1

    Steps

    1. exe^{x} integrates to exe^{x}; subtract the lower end from the upper
    2. e1−e0\displaystyle e^{1}-e^{0}=e−1\displaystyle = e-1
    3. We subtract e0=1e^{0}=1, so it is not ee itself
  5. 05What is the value?e−(e−1)\displaystyle e - (e-1)AnswerHide
    Answer1\displaystyle 1

    Steps

    1. Removing the brackets makes the two ee cancel
    2. e−(e−1)\displaystyle e-(e-1)=e−e+1\displaystyle = e-e+1=1\displaystyle = 1
    3. The minus in front flips the −1-1 inside into +1+1
  6. 06What is the value?[x2]02\displaystyle [x^{2}]_{0}^{2}AnswerHide
    Answer4\displaystyle 4

    Steps

    1. Put in the upper limit, then subtract the value at the lower limit
    2. 22−02\displaystyle 2^{2} - 0^{2}=4\displaystyle = 4

FAQ

Q1Do I add a constant of integration when finding vv?AnswerHide
A
No need. Any one antiderivative of v′v' works; for v′=cos⁡xv'=\cos x, take v=sin⁡xv=\sin x.
Q2Does the integral being subtracted keep the same limits?AnswerHide
A
Yes. Both [uv]ab\left[uv\right]_{a}^{b} and ∫abu′v dx\int_{a}^{b} u'v\,dx use the same limits aa to bb. Remember that the minus sign applies to the whole second integral.

Published 2026-10-14 · Updated 2026-10-14