Integration by parts for a useful product, integrals from easy to advanced lesson

The integral $\int_0^1 xe^x\,dx$ multiplies a linear factor by an exponential. Substitution does not finish the job here. The choice $u = e^x$ , for instance, only turns it into $\int_1^e \ln u\,du$ , which still needs an antiderivative of the logarithm. The two factors have different strengths instead, since differentiating $x$ gives the constant $1$ and integrating $e^x$ gives $e^x$ again. This page derives integration by parts from the product rule, uses it to show that the integral equals $1$ , and reads the same number as the area of a region above the exponential curve.

Key terms

Product rule in reverse

Definition If $u$ and $v$ are differentiable on an interval, the product rule says $(uv)' = u'v + uv'$ there, so $uv$ is an antiderivative of $u'v + uv'$ . If $u$ and $v$ are continuously differentiable on $[a, b]$ , the fundamental theorem of calculus then gives

\int_a^b \bigl(u'v + uv'\bigr)\,dx = \bigl[uv\bigr]_a^b.

In plain words A product changes for two reasons, because the first factor changes and because the second factor changes. Adding up both kinds of change across an interval gives the total change of the product between the endpoints.

Example For $u = x$ and $v = e^x$ the product rule gives $(xe^x)' = e^x + xe^x$ , so

\begin{aligned} \int_0^1 \bigl(e^x + xe^x\bigr)\,dx &= \bigl[xe^x\bigr]_0^1 \\[0.4em] &= 1\cdot e - 0\cdot 1 = e. \end{aligned}

The first piece alone is $\int_0^1 e^x\,dx = e - 1$ , so the second piece must be $\int_0^1 xe^x\,dx = e - (e - 1) = 1$ . That is integration by parts in miniature.

Integration by parts

Definition Let $u$ and $v$ be continuously differentiable on $[a, b]$ . Then

\begin{aligned} \int_a^b u\,v'\,dx &= \bigl[uv\bigr]_a^b \\[0.4em] &\quad - \int_a^b u'\,v\,dx. \end{aligned}

It follows from the product rule in reverse by splitting the integral of $u'v + uv'$ into two and moving $\int_a^b u'v\,dx$ to the other side. On an interval where $u$ and $v$ are continuously differentiable, the indefinite form is $\int u\,v'\,dx = uv - \int u'\,v\,dx$ , usually written $\int u\,dv = uv - \int v\,du$ with $du = u'(x)\,dx$ and $dv = v'(x)\,dx$ .

In plain words Pick one factor to differentiate and one to integrate. The integral of the product equals the product evaluated at the ends, minus a new integral in which the first factor has been differentiated and the second integrated. The method pays off when that new integral is easier than the old one.

Example For $\int_0^1 xe^x\,dx$ take $u = x$ and $dv = e^x\,dx$ , so $du = dx$ and $v = e^x$ . Both $x$ and $e^x$ are continuously differentiable on $[0, 1]$ . Then

\begin{aligned} \int_0^1 xe^x\,dx &= \bigl[xe^x\bigr]_0^1 \\[0.4em] &\quad - \int_0^1 e^x\,dx \\[0.4em] &= e - (e - 1) = 1. \end{aligned}

The antiderivative $F(x) = e^x(x - 1)$ confirms the value, since $F'(x) = e^x(x - 1) + e^x = xe^x$ and $F(1) - F(0) = 0 - (-1) = 1$ . This antiderivative is negative on $[0, 1)$ even though the integral is positive, because only the difference $F(1) - F(0)$ matters. The size is plausible as well. For $0 \le x \le 1$ we have $1 \le e^x \le e$ , so $x \le xe^x \le ex$ , and the integral lies between $\tfrac12$ and $\tfrac{e}{2} \approx 1.359$ .

Boundary term

Definition The boundary term of integration by parts on $[a, b]$ is

\begin{aligned} \bigl[uv\bigr]_a^b &= u(b)\,v(b) \\[0.4em] &\quad - u(a)\,v(a), \end{aligned}

the change of the product $uv$ between the endpoints. In the indefinite form its counterpart is the product $uv$ itself. It is part of the formula, so it is never optional, and each of its two values is a product of both factors at the same endpoint.

In plain words Before working on the new integral, evaluate $u$ times $v$ at the upper end and at the lower end and subtract. Leaving the term out, or evaluating only one factor at an endpoint, changes the answer.

Example For $u = x$ and $v = e^x$ on $[0, 1]$ ,

\bigl[xe^x\bigr]_0^1 = 1\cdot e - 0\cdot 1 = e.

The lower value is $0$ because the factor $x$ is $0$ there, not because the exponential vanishes, since $e^0 = 1$ . Leaving out the boundary term would give $-(e - 1) \approx -1.718$ , a negative value for the integral of a function that is nonnegative on $[0, 1]$ , which is impossible.

Choosing u and dv

Definition Integration by parts needs a split of the integrand into a factor $u$ , which will be differentiated, and a factor $dv = v'(x)\,dx$ , whose antiderivative $v$ must be found. Every split with $u$ and $v$ continuously differentiable gives a true equation. The choice only decides whether the new integral $\int u'\,v\,dx$ is easier than the original. A common rule of thumb ranks the types logarithmic, inverse trigonometric, algebraic, trigonometric and exponential in that order, and takes $u$ from the earliest type present. It is a guide, not a theorem.

In plain words Differentiate the factor that gets simpler when differentiated, such as a power of $x$ or a logarithm. Integrate the factor that stays manageable when integrated, such as $e^x$ , $\sin x$ or $\cos x$ .

Example With $u = x$ and $dv = e^x\,dx$ the new integrand is $1\cdot e^x$ , and the factor $x$ is gone. The reverse choice $u = e^x$ and $dv = x\,dx$ gives $v = \tfrac12 x^2$ and

\begin{aligned} \int_0^1 xe^x\,dx &= \Bigl[\frac{x^2}{2}e^x\Bigr]_0^1 \\[0.4em] &\quad - \frac12\int_0^1 x^2e^x\,dx. \end{aligned}

This equation is true, since $\int_0^1 x^2e^x\,dx = e - 2$ and $\tfrac{e}{2} - \tfrac12(e - 2) = 1$ , but the new integral has a higher power of $x$ than the original and is harder to evaluate.

Complementary area

Definition Let $f$ be continuously differentiable and strictly increasing on $[0, b]$ . The complementary area is the area of the region between the graph of $f$ and the horizontal line $y = f(b)$ , the points with $0 \le x \le b$ and $f(x) \le y \le f(b)$ . Vertical strips give it as $bf(b) - \int_0^b f(x)\,dx$ . Horizontal strips at height $y$ run from $x = 0$ to $x = f^{-1}(y)$ and give it as $\int_{f(0)}^{f(b)} f^{-1}(y)\,dy$ . Integration by parts with $u = x$ and $v = f$ , and the substitution $y = f(x)$ , connect both to $\int_0^b x f'(x)\,dx$ ,

\begin{aligned} \int_0^b x f'(x)\,dx &= bf(b) \\[0.4em] &\quad - \int_0^b f(x)\,dx \\[0.4em] &= \int_{f(0)}^{f(b)} f^{-1}(y)\,dy. \end{aligned}

In plain words When $f(0) \ge 0$ , the rectangle of width $b$ and height $f(b)$ splits into the region under the curve and the region above it. Integration by parts computes the region above the curve as an integral in $x$ . Read with horizontal strips, the width of that region at height $y$ is the position $x$ where the curve reaches height $y$ .

Example For $f(x) = e^x$ on $[0, 1]$ the rectangle has area $e$ and the region under the curve has area $e - 1$ , so the region above the curve and inside the rectangle has area $1$ . At a height $y$ between $1$ and $e$ the curve is reached at $x = \ln y$ , so horizontal strips give

\begin{aligned} \int_1^e \ln y\,dy &= \bigl[y\ln y - y\bigr]_1^e \\[0.4em] &= 0 - (-1) = 1. \end{aligned}

The substitution $y = e^x$ , with $dy = e^x\,dx$ , turns this integral into $\int_0^1 xe^x\,dx$ , because each horizontal strip has width $x$ and height $dy = e^x\,dx$ . The region under $y = xe^x$ and the region above $y = e^x$ have different shapes, but both have area $1$ .

Check yourself

Evaluate the integral

\int_0^\pi x\sin x\,dx.

Show the answer

Take $u = x$ and $dv = \sin x\,dx$ , so $du = dx$ and $v = -\cos x$ . Then

\begin{aligned} \int_0^\pi x\sin x\,dx &= \bigl[-x\cos x\bigr]_0^\pi \\[0.4em] &\quad + \int_0^\pi \cos x\,dx \\[0.4em] &= \pi + 0 = \pi. \end{aligned}

The boundary term is $-\pi\cos\pi - (-0\cdot\cos 0) = \pi$ , and $\int_0^\pi \cos x\,dx = \sin\pi - \sin 0 = 0$ . Differentiating the antiderivative $\sin x - x\cos x$ gives $\cos x - \cos x + x\sin x = x\sin x$ , which confirms it.

For the integral below, which factor should be $u$ ? Evaluate it.

\int_1^e x\ln x\,dx

Show the answer

Take $u = \ln x$ and $dv = x\,dx$ , so $du = \tfrac1x\,dx$ and $v = \tfrac12 x^2$ , and $u$ and $v$ are continuously differentiable on $[1, e]$ . Differentiating the logarithm leaves only a power of $x$ . Then

\begin{aligned} \int_1^e x\ln x\,dx &= \Bigl[\frac{x^2}{2}\ln x\Bigr]_1^e \\[0.4em] &\quad - \int_1^e \frac{x}{2}\,dx \\[0.4em] &= \frac{e^2}{2} - \frac{e^2 - 1}{4} = \frac{e^2 + 1}{4}. \end{aligned}

The value is about $2.097$ . Taking $u = x$ out of habit is a detour here, because then $dv = \ln x\,dx$ and finding $v = x\ln x - x$ already needs integration by parts, and the new integral $\int (x\ln x - x)\,dx$ contains the original one again. The size checks out, since $0 \le x\ln x \le x$ on $[1, e]$ gives a value between $0$ and $\tfrac12(e^2 - 1) \approx 3.19$ .

Use $\int_0^1 xe^x\,dx = 1$ to evaluate the integral below. Why is $-2$ , the value obtained by leaving out the boundary term, impossible?

\int_0^1 x^2e^x\,dx

Show the answer

Take $u = x^2$ and $dv = e^x\,dx$ , so $du = 2x\,dx$ and $v = e^x$ . Then

\begin{aligned} \int_0^1 x^2e^x\,dx &= \bigl[x^2e^x\bigr]_0^1 \\[0.4em] &\quad - 2\int_0^1 xe^x\,dx \\[0.4em] &= e - 2\cdot 1 = e - 2. \end{aligned}

The value is about $0.718$ . The integrand is nonnegative on $[0, 1]$ , and since $x^2 \le x^2e^x \le ex^2$ there, the integral lies between $\tfrac13$ and $\tfrac{e}{3} \approx 0.906$ . A negative value such as $-2$ cannot be the integral of a nonnegative function, so the boundary term $e$ cannot be left out.

Integration by parts for a useful product

Key terms

Product rule in reverse

Integration by parts

Boundary term

Choosing u and dv

Complementary area

Check yourself

How this lesson connects

Builds on