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Calculus Textbooks Boundless Calculus Inverse Functions and Advanced Integration Techniques of Integration
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Calculus
Concept Version 7
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Integration By Parts

Integration by parts is a way of integrating complex functions by breaking them down into separate parts and integrating them individually.

Learning Objective

  • Solve integrals by using integration by parts


Key Points

    • Integration by parts is a theorem that relates the integral of a product of functions to the integral of their derivative and anti-derivative.
    • The theorem is expressed as $\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$.
    • Integration by parts may be interpreted graphically in addition to mathematically.

Terms

  • integral

    also sometimes called antiderivative; the limit of the sums computed in a process in which the domain of a function is divided into small subsets and a possibly nominal value of the function on each subset is multiplied by the measure of that subset, all these products then being summed

  • derivative

    a measure of how a function changes as its input changes


Full Text

Introduction

In calculus, integration by parts is a theorem that relates the integral of a product of functions to the integral of their derivative and anti-derivative. It is frequently used to find the anti-derivative of a product of functions into an ideally simpler anti-derivative. The rule can be derived in one line by simply integrating the product rule of differentiation.

Theorem of integration by parts

Let's take the functions $u = u(x)$ and $v = v(x)$. When taking their derivatives, we are left with $du = u '(x)$ and $dxdv = v'(x) dx$. Now, let's take a look at the principle of integration by parts:

$\displaystyle{\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \ dx}$

or, more compactly,

$\displaystyle{\int u \, dv=uv-\int v \, du}$

Proof

Suppose $u(x)$ and $v(x)$ are two continuously differentiable functions. The product rule states: 

$\displaystyle{\frac{d}{dx}\left(u(x)v(x)\right) = u(x) \frac{d}{dx}\left(v(x)\right) + \frac{d}{dx}\left(u(x)\right) v(x)}$

Integrating both sides with respect to $x$, over an interval $a \leq x \leq b$,

$\displaystyle{\int_a^b \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx}$ 

then applying the fundamental theorem of calculus,

$\displaystyle{\int_a^b \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \left[u(x)v(x)\right]_a^b}$

 gives the formula for "integration by parts":

$\displaystyle{\left[u(x)v(x)\right]_a^b = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx}$.

Visulization

Let's define a parametric curve by $(x, y) = (f(t), g(t))$.

Integration By Parts

Integration by parts may be thought of as deriving the area of the blue region from the total area and that of the red region. The area of the blue region is $A_1=\int_{y_1}^{y_2}x(y)dy$. Similarly, the area of the red region is $A_2=\int_{x_1}^{x_2}y(x)dx$. The total area, $A_1+A_2$, is equal to the area of the bigger rectangle, $x_2y_2$, minus the area of the smaller one, $x_1y_1$: $\int_{y_1}^{y_2}x(y)dy+\int_{x_1}^{x_2}y(x)dx=\biggl.x_iy_i\biggl|_{i=1}^{i=2}$. Assuming the curve is smooth within a neighborhood, this generalizes to indefinite integrals $\int xdy + \int y dx = xy$, which can be rearranged to the form of the theorem: $\int xdy = xy - \int y dx$.

Example

In order to calculate $I=\int x\cos (x) \,dx$, let:

$u = x \\ \therefore du = dx$

and

$dv = \cos(x)\,dx \\ \therefore v = \int\cos(x)\,dx = \sin x$

then:

$\begin{aligned} \int x\cos (x) \,dx & = \int u \, dv \\ & = uv - \int v \, du \\ & = x\sin (x) - \int \sin (x) \,dx \\ & = x\sin (x) + \cos (x) + C \end{aligned}$

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