Title: Default for 13.4 Integration by Parts
Course: HELM Workbooks 2023-2024

Integration by parts

Integration by Parts is a technique for integrating products of functions. In this Section you will learn to recognise when it is appropriate to use the technique and have the opportunity to practise using it for finding both definite and indefinite integrals.

Indefinite integration

The technique known as integration by parts is used to integrate a product of two functions, such as in these two examples:

- \({\displaystyle\int e^{2x}\sin 3x \, \mathrm{d}x}\)
- \({\displaystyle\int_0^1 x^3 e^{-2x} \, \mathrm{d}x}\)

Note that in the first example, the integrand is the product of the functions \(e^{2x}\) and \(\sin 3x\), and in the second example the integrand is the product of the functions \(x^3\) and \(e^{-2x}\). Note also that we can change the order of the terms in the product if we wish and write

- \({\displaystyle\int (\sin 3x) e^{2x} \, \mathrm{d}x}\)
- \({\displaystyle\int_0^1 e^{-2x}x^3 \, \mathrm{d}x}\)

What you must never do is integrate each term in the product separately and then multiply - the integral of a product is not the product of the separate integrals. However, it is often possible to find integrals involving products using the method of integration by parts - you can think of this as a product rule for integrals.

The integration by parts formula states:

Key Point

Integration by Parts for Indefinite Integrals

given functions \(f(x)\) and \(g(x)\):

\({\displaystyle\int f \cdot g \, \mathrm{d}x}\) = \(f\) \(\cdot\) \({\displaystyle\int g \, \mathrm{d}x}\) - \({\displaystyle\int \left(\dfrac{df}{dx} \displaystyle\int g \, \mathrm{d}x \right) \, \mathrm{d}x}\)

Alternatively, given functions \(u\) and \(v\):

\({\displaystyle \int u\dfrac{d v}{d x} \, \mathrm{d}x = u.v-\displaystyle \int v\dfrac{d u}{d x} \, \mathrm{d}x}\)

Study the formula carefully and note the following observations. Firstly, to apply the formula we must be able to differentiate the function \(f\) to find \(\dfrac{d f}{d x}\), and we must be able to integrate the function, \(g\).

Secondly the formula replaces one integral, the one on the left, with a different integral, that on the far right. The intention is that the latter, whilst it may look more complicated in the formula above, is simpler to evaluate. Consider the following Example:

Example

Find the integral of the product of \(x\) with \(\sin x\); that is, find \({\displaystyle \int x\sin x \, \mathrm{d}x}\).

Solution

Compare the required integral with the formula for integration by parts: we choose \(f = x\) and \(g = \sin x\)

It follows that \(\dfrac{df}{dx} = 1\) and \({\displaystyle\int g \, \mathrm{d}x = \displaystyle\int \sin x \, \mathrm{d}x} = -\cos x\)

(When integrating \(g\) there is no need to worry about a constant of integration. When you become confident with the method, you may like to think about why this is the case.)

Applying the formula we obtain

\begin{eqnarray*} \int x\sin x \, \mathrm{d} x &=& f\cdot \int g \, \mathrm{d} x -\int \left( {d f\over d x}\cdot\int g \, \mathrm{d} x \right)\,d x \\ &=& x(-\cos x) - \int 1(-\cos x) \, \mathrm{d} x\\ &=& -x\cos x + \int \cos x \, \mathrm{d} x\\ &=& -x\cos x + \sin x +c \end{eqnarray*}

Exercise

In some questions below it will be necessary to apply integration by parts more than once.

Find:

1. \({\displaystyle\int [value] \, \mathrm{d}x} = \)
2. \({\displaystyle\int [value] \, \mathrm{d}t} = \)
3. \({\displaystyle\int x\cos x \, \mathrm{d}x} = \)
Feedback: We apply integraton by parts: \[ \int u \frac{dv}{dx} \, \mathrm{d}x = uv- \int \frac{du}{dx} v \, \mathrm{d}x \] or alternatively: \[ \int f(x)g(x) \, \mathrm{d}x = f(x) \, \int g(x) \, \mathrm{d}x - \int \frac{df}{dx} \left\{ \int g(x) \, \mathrm{d}x \right\} \, \mathrm{d}x. \] 1. We are given the integral \({\displaystyle\int [value] \, \mathrm{d}x}.\) We have \( f(x) = [value]\) and \(g(x) = [value].\) Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=[value] \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value]\mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x \\ &= [value]. \end{aligned}\] 2. We are given the integral \({\displaystyle\int [value] \, \mathrm{d}x}.\) We have \( f(x) = [value]\) and \(g(x) = [value].\) Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=[value] \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value] \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x \\ &= [value]. \end{aligned}\] 3. We are given the integral \({\displaystyle\int [value] \, \mathrm{d}x}.\) We have \( f(x) = [value]\) and \(g(x) = [value].\) Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=[value] \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value] \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x \\ &= [value]. \end{aligned}\]

Definite integration

When dealing with definite integrals the relevant formula is as follows:

Key Point

Integration by Parts for Definite Integrals

For definite integrals, given functions \(f(x)\) and \(g(x)\): \[{\displaystyle\int_a^b f \cdot g \, \mathrm{d}x} = \left[f \cdot {\displaystyle\int g \, \mathrm{d}x}\right]_a^b - {\displaystyle\int_a^b \left(\dfrac{\mathrm{d}f}{\mathrm{d}x} \displaystyle\int g \, \mathrm{d}x \right) \, \mathrm{d}x}\]

Alternatively, given functions \(u\) and \(v\): \[{\displaystyle \int_a^b u\dfrac{\mathrm{d} v}{\mathrm{d} x} \,\mathrm{d}x = \left[u \cdot v\right]_a^b-\displaystyle \int_a^b v\dfrac{\mathrm{d} u}{\mathrm{d} x} \, \mathrm{d}x}\]

Example

Find \({\displaystyle\int_0^2 x e^x \, \mathrm{d}x}\)

Solution

We let \(f = x\) and \(g = e^x\).

Then \(\dfrac{\mathrm{d}f}{\mathrm{d}x} = 1\) and \({\displaystyle\int g \, \mathrm{d}x} = e^x\).

Using integration by parts we obtain: \[\begin{aligned}\int_0^2 x e^x \, \mathrm{d}x &= \left[x e^x\right]_0^2 - \int_0^2 1 \cdot e^x \, \mathrm{d}x \\ &= 2 e^2 - \left[e^x\right]_0^2 \\ &= 2 e^2 - [e^2 - 1] = e^2 +1 \end{aligned}\]

(or \(8.389\) to 3 d.p.)

Sometimes it is necessary to apply the formula more than once as the next Example shows.

Example

Find \({\displaystyle\int_0^2 x^2 e^x \, \mathrm{d}x}\)

Solution

We let \(f = x^2\) and \(g = e^x\).

Then \(\dfrac{\mathrm{d}f}{\mathrm{d}x} = 2x\) and \({\displaystyle\int g \, \mathrm{d}x} = e^x\).

Using integration by parts we obtain: \[\begin{aligned}\int_0^2 x^2 e^x \, \mathrm{d}x &= \left[x^2 e^x\right]_0^2 - \int_0^2 2x e^x \, \mathrm{d}x \\ &= 4e^2 - 2\int_0^2 x e^x \, \mathrm{d}x \end{aligned}\]

The remaining integral must be integrated by parts also but we have just done this in the example above. So \[ \int_0^2 x^2 e^x \, \mathrm{d}x = 4e^2- 2[e^2+1] = 2e^2-2 \]

(or \(12.778\) to 3 d.p.)

Exercise

Evaluate the following. Answers should be given as a decimal rounded to \(4\) d.p.
1. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x} = \)
2. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x} = \)
3. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}t} = \)
Feedback: [[facts:calc_int_methods_parts]] 1. First we find \(\displaystyle \frac{d}{dx}\left([value]\right)\) and \(\displaystyle \int [value] \, \mathrm{d}x\). \[\begin{aligned} \frac{d}{dx}\left([value]\right) &= [value] \\ \int [value]\,\mathrm{d}x &= [value] \end{aligned}\] The formula then gives \[\begin{aligned} \int_{[value]}^{[value]} [value] \,\mathrm{d}x &= \Bigg[\left([value]\right) \cdot \left([value]\right)\Bigg]_{[value]}^{[value]} - \int_{[value]}^{[value]} \left([value]\right) \cdot \left([value]\right) \,\mathrm{d}x \\ &= \Bigg[[value]\Bigg]_{[value]}^{[value]} - \Bigg[[value]\Bigg]_{[value]}^{[value]} \\ &= \left([value] - \left([value]\right)\right) - \left([value] - \left([value]\right)\right) \\ &= [value]. \end{aligned}\] 2. First we find \(\displaystyle \frac{d}{dx}\left([value]\right)\) and \(\displaystyle \int [value] \, \mathrm{d}x\). \[\begin{aligned} \frac{d}{dx}\left([value]\right) &= [value] \\ \int [value]\,\mathrm{d}x &= [value] \end{aligned}\] The formula then gives \[\begin{aligned} \int_{[value]}^{[value]} [value] \,\mathrm{d}x &= \Bigg[\left([value]\right) \cdot \left([value]\right)\Bigg]_{[value]}^{[value]} - \int_{[value]}^{[value]} \left([value]\right) \cdot \left([value]\right) \,\mathrm{d}x \\ &= \Bigg[[value]\Bigg]_{[value]}^{[value]} - \Bigg[[value]\Bigg]_{[value]}^{[value]} \\ &= \left([value] - \left([value]\right)\right) - \left([value] - \left([value]\right)\right) \\ &= [value]. \end{aligned}\] 3. First we find \(\displaystyle \frac{d}{dt}\left([value]\right)\) and \(\displaystyle \int [value] \, \mathrm{d}t\). \[\begin{aligned} \frac{d}{dt}\left([value]\right) &= [value] \\ \int [value]\,\mathrm{d}t &= [value] \end{aligned}\] The formula then gives \[\begin{aligned} \int_{[value]}^{[value]} [value] \,\mathrm{d}t &= \Bigg[\left([value]\right) \cdot \left([value]\right)\Bigg]_{[value]}^{[value]} - \int_{[value]}^{[value]} \left([value]\right) \cdot \left([value]\right) \,\mathrm{d}t \\ &= \Bigg[[value]\Bigg]_{[value]}^{[value]} - \Bigg[[value]\Bigg]_{[value]}^{[value]} \\ &= \left([value] - \left([value]\right)\right) - \left([value] - \left([value]\right)\right) \\ &= [value]. \end{aligned}\]

Exercise

Find

\({\displaystyle\int [value] \, \mathrm{d}x} = \)
Feedback: We apply integraton by parts: \[ \int u \frac{dv}{dx} \, \mathrm{d}x = uv - \int \frac{du}{dx} v \, \mathrm{d}x, \] or alternatively: \[\int f(x)g(x) \, \mathrm{d}x = f(x) \, \int g(x) \mathrm{d}x - \int \frac{df}{dx} \left\{\int g(x) \, \mathrm{d}x \right\} \, \mathrm{d}x. \] We are given the integral \({\displaystyle\int [value] \, \mathrm{d}x}\). We have \( f(x) = [value]\) and \(g(x) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=([value]) \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value] \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x \\ &= [value]. \end{aligned}\]

Exercise

By writing \(\ln x\) as \(1\times \ln x\) find

\(\displaystyle \int \ln x\, \mathrm{d}x = \)
Feedback: We apply integraton by parts: \[ \int u \frac{dv}{dx} \, \mathrm{d}x = uv - \int \frac{du}{dx} v \, \mathrm{d}x, \] or alternatively: \[\int f(x)g(x) \, \mathrm{d}x = f(x) \, \int g(x) \mathrm{d}x - \int \frac{df}{dx} \left\{\int g(x) \, \mathrm{d}x \right\} \, \mathrm{d}x. \] We are given the integral \(\displaystyle\int [value] \, \mathrm{d}x\). We have \( f(x) = [value]\) and \(g(x) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=[value] \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value] \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x \\ &= [value]. \end{aligned}\]

Exercise

In some questions below it will be necessary to apply integration by parts more than once.

Find:

1. \({\displaystyle\int \tan^{-1}(x) \, \mathrm{d}x} =\)
2. \({\displaystyle \int [value] \, \mathrm{d}x} =\)
3. \({\displaystyle \int [value] \, \mathrm{d}x} =\)
Feedback: We apply integraton by parts: \[ \int u \frac{dv}{dx} \, \mathrm{d}x = uv - \int \frac{du}{dx} v \, \mathrm{d}x, \] or alternatively: \[\int f(x)g(x) \, \mathrm{d}x = f(x) \, \int g(x) \mathrm{d}x - \int \frac{df}{dx} \left\{\int g(x) \, \mathrm{d}x \right\} \, \mathrm{d}x. \] 1. Writing \(\displaystyle{\int \tan^{-1}x \, \mathrm{d}x =\int 1 \cdot \tan^{-1}x \, \mathrm{d}x}\) we have \(f(x)=\tan^{-1}\) and \(g(x)=1\). Hence: \[\begin{aligned} \int \tan^{-1}(x) \, \mathrm{d}x &=\tan^{-1}(x) \int 1 \, \mathrm{d}x -\int \left( \dfrac{d}{dx}\tan^{-1}(x)\right) \cdot\left(\int 1 \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= x\tan^{-1}(x)- \int \dfrac{x}{1+x^2} \, \mathrm{d}x \\ &= [value]. \end{aligned}\] 2. We are given the integral \({\displaystyle\int [value] \, \mathrm{d}x}\). We have \( f(x) = [value]\) and \(g(x) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=[value] \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value] \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x \\ &= [value]. \end{aligned}\] 3. We are given the integral \(\displaystyle{\int [value] \, \mathrm{d}x}\). We have \( f(x) = [value]\) and \(g(x) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=[value] \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value] \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x. \end{aligned}\] We need to find \(\displaystyle \int [value] \, \mathrm{d}x\) using integration by parts again. We get: \[ \int [value] \, \mathrm{d}x = [value].\] Altogether we have: \[\int [value] \mathrm{d}x = [value]. \]

Exercise

In some questions below it will be necessary to apply integration by parts more than once.

Find:

1. \({\displaystyle\int x\cos kx \, \mathrm{d}x}\), where \(k\) is a constant:
2. \({\displaystyle\int z^2\cos kz \, \mathrm{d}z}\), where \(k\) is a constant:
Feedback: We apply integraton by parts: \[ \int u \frac{dv}{dx} \, \mathrm{d}x = uv - \int \frac{du}{dx} v \, \mathrm{d}x, \] or alternatively: \[\int f(x)g(x) \, \mathrm{d}x = f(x) \, \int g(x) \mathrm{d}x - \int \frac{df}{dx} \left\{\int g(x) \, \mathrm{d}x \right\} \, \mathrm{d}x. \] 1. We are given the integral \({\displaystyle\int [value] \, \mathrm{d}x}\). We have \( f(x) = [value]\) and \(g(x) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}x &=[value] \int [value] \, \mathrm{d}x -\int \left([value] \cdot \int [value] \, \mathrm{d}x \right) \, \mathrm{d}x \\ &= [value]- \int [value] \, \mathrm{d}x \\ &= [value]. \end{aligned}\] 2. We are given the integral \(\displaystyle{ \int [value] \, \mathrm{d}z}\). We have \( f(z) = [value]\) and \(g(z) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}z &=[value] \int [value] \, \mathrm{d}z -\int \left([value] \cdot \int [value] \, \mathrm{d}z \right) \, \mathrm{d}z \\ &= [value]- \int [value] \, \mathrm{d}z \end{aligned}\] We now need to find \(\displaystyle \int [value] \, \mathrm{d}z\) using integration by part again. We get: \[\int [value] \, \mathrm{d}z = [value]\] Altogether we have: \[\displaystyle{ \int [value] \, \mathrm{d}z= [value]}.\]

Exercise

In some questions below it will be necessary to apply integration by parts more than once.

Find:

1. \({\displaystyle\int t e^{-st} \, \mathrm{d}t}\) where \(s\) is a constant:
2. \({\displaystyle\int t^2 e^{-st} \, \mathrm{d}t}\) where \(s\) is a constant:
Feedback: We apply integraton by parts: \[ \int u \frac{dv}{dx} \, \mathrm{d}x = uv - \int \frac{du}{dx} v \, \mathrm{d}x, \] or alternatively: \[\int f(x)g(x) \, \mathrm{d}x = f(x) \, \int g(x) \mathrm{d}x - \int \frac{df}{dx} \left\{\int g(x) \, \mathrm{d}x \right\} \, \mathrm{d}x. \] 1. We are given the integral \({\displaystyle\int [value] \, \mathrm{d}t}\). We have \( f(x) = [value]\) and \(g(x) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}t &=[value] \int [value] \, \mathrm{d}t-\int \left([value] \cdot \int [value] \, \mathrm{d}t \right) \, \mathrm{d}t \\ &= [value]- \int [value] \, \mathrm{d}t \\ &= [value]. \end{aligned}\] 2. We are given the integral \({\displaystyle\int [value] \, \mathrm{d}t}\). We have \( f(x) = [value]\) and \(g(x) = [value]\). Hence: \[\begin{aligned} \int [value] \, \mathrm{d}t &=[value] \int [value] \, \mathrm{d}t-\int \left([value] \cdot \int [value] \, \mathrm{d}t \right) \, \mathrm{d}t \\ &= [value]- \int [value] \, \mathrm{d}t. \end{aligned}\] We now need to find \(\displaystyle \int [value] \, \mathrm{d}t \) using integration by parts again. We get: \[\displaystyle {\int [value] \, \mathrm{d}t= [value]}.\] Altogether we have: \[\displaystyle{\int [value] \, \mathrm{d}t= [value]}.\]

Exercise

Find the values of the following integrals giving your answers to \(4\) d.p.

1. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x} = \)
2. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x} = \)
Feedback: [[facts:calc_int_methods_parts]] 1. First we find \(\displaystyle \frac{d}{dx}\left([value]\right)\) and \(\displaystyle \int [value] \, \mathrm{d}x\). \[\begin{aligned} \frac{d}{dx}\left([value]\right) &= [value] \\ \int [value]\,\mathrm{d}x &= [value] \end{aligned}\] The formula then gives \[\begin{aligned} \int_{[value]}^{[value]} [value] \,\mathrm{d}x &= \Bigg[\left([value]\right) \cdot \left([value]\right)\Bigg]_{[value]}^{[value]} - \int_{[value]}^{[value]} \left([value]\right) \cdot \left([value]\right) \,\mathrm{d}x \\ &= \Bigg[[value]\Bigg]_{[value]}^{[value]} - \Bigg[[value]\Bigg]_{[value]}^{[value]} \\ &= \left([value] - \left([value]\right)\right) - \left([value] - \left([value]\right)\right) \\ &= [value]. \end{aligned}\] 2. First we find \(\displaystyle \frac{d}{dx}\left([value]\right)\) and \(\displaystyle \int [value] \, \mathrm{d}x\). \[\begin{aligned} \frac{d}{dx}\left([value]\right) &= [value] \\ \int [value]\,\mathrm{d}x &= [value] \end{aligned}\] The formula then gives \[\begin{aligned} \int_{[value]}^{[value]} [value] \,\mathrm{d}x &= \Bigg[\left([value]\right) \cdot \left([value]\right)\Bigg]_{[value]}^{[value]} - \int_{[value]}^{[value]} \left([value]\right) \cdot \left([value]\right) \,\mathrm{d}x \\ &= \Bigg[[value]\Bigg]_{[value]}^{[value]} - \Bigg[[value]\Bigg]_{[value]}^{[value]} \\ &= \left([value] - \left([value]\right)\right) - \left([value] - \left([value]\right)\right) \\ &= [value]. \end{aligned}\]

Exercise

Find \(\displaystyle \int [value] \, \mathrm{d}x\).

Let \(f = [value]\) and \(g = [value]\). Now calculate \(\displaystyle \frac{df}{dx}\) and \(\displaystyle \int g \, \mathrm{d}x\). You do not need to add the constant of integration in the second case.

\(\displaystyle \frac{df}{dx} = \)

\(\displaystyle \int g \, \mathrm{d}x = \)

Substitute these results into the formula for integration by parts and complete the question. Make sure to include the constant of integration in your final answer.

\(\displaystyle \int [value] \, \mathrm{d}x = \)
Feedback: We are asked to find \(\displaystyle \int [value] \, \mathrm{d}x\). In order to do this, we will use integration by parts. Here \(f = [value]\) and \(g = [value]\). First we find \(\displaystyle \frac{df}{dx}\) and \(\displaystyle \int g \, \mathrm{d}x\). \[\begin{aligned} \frac{df}{dx} &= [value] \\ \int g \, \mathrm{d}x &= [value] \end{aligned}\] Now we use the formula for integration by parts: \[\int f \cdot g \, \mathrm{d}x = f \cdot \int g \, \mathrm{d}x - \int \left(\dfrac{df}{dx} \int g \, \mathrm{d}x \right) \, \mathrm{d}x\] This gives \[\begin{aligned} \int [value] \, \mathrm{d}x &= \left([value]\right) \cdot \left([value]\right) - \int \left([value]\right) \cdot \left([value]\right) \, \mathrm{d}x \\ &= [value] - \int [value] \, \mathrm{d}x \\ &= [value] - \left([value]\right) \\ &= [value]. \end{aligned}\]

Sometimes it is necessary to apply the formula more than once, as the next Example shows.

Example

Find \({\displaystyle\int 2x^2 e^{-x} \, \mathrm{d}x}\).

Solution

We let \(f=2x^2\) and \(g=e^{-x}\). Then \(\dfrac{\mathrm{d} f}{\mathrm{d} x}=4x\) and \({\displaystyle\int g \, \mathrm{d}x} = -e^{-x}\)

Using the formula for integration by parts we find

\({\displaystyle\int 2x^2 e^{-x} \, \mathrm{d}x = 2x^2(-e^{-x})-\displaystyle\int 4x(-e^{-x}) \, \mathrm{d}x = -2x^2e^{-x}+\displaystyle\int4xe^{-x} \, \mathrm{d}x}\)

We now need to find \({\displaystyle\int4xe^{-x} \, \mathrm{d}x}\) using integration by parts again. We get

\begin{eqnarray*} \int 4xe^{-x}\ \mathrm{d} x&=&4x(-e^{-x})-\int4(-e^{-x})\ \mathrm{d} x\\ &=&-4xe^{-x}+\int 4e^{-x}\ \mathrm{d} x\\ &=&-4xe^{-x}-4e^{-x}\end{eqnarray*}

Altogether we have

\({\displaystyle\int 2x^2e^{-x} \, \mathrm{d}x} = -2x^2e^{-x}-4x e^{-x}-4 e^{-x}+c = -2e^{-x}(x^2+2x+2)+c\)

Exercise

Find \(\displaystyle \int_0^{[value]} [value] \, \mathrm{d}x\) to \(3\) d.p.

What are your choices for \(f\) and \(g\)?

\(f(x) = \)

\(g(x) = \)

Now complete the integral.

\(\displaystyle \int_0^{[value]} [value] \, \mathrm{d}x = \)
Feedback: We are asked to find \(\displaystyle \int_0^{[value]} [value] \, \mathrm{d}x\). In order to do this we use integration by parts. We must choose \(f(x) = [value]\) and \(g(x) = [value]\). Now we use the integration by parts formula: \[\int_a^b f \cdot g \, \mathrm{d}x = \left[f \cdot \int g \, \mathrm{d}x \right]_a^b - \int_a^b \left(\dfrac{df}{dx} \int g \, \mathrm{d}x \right) \, \mathrm{d}x\] First we find \(\displaystyle \frac{df}{dx}\) and \(\displaystyle \int g(x) \, \mathrm{d}x\). \[\begin{aligned} \frac{df}{dx} &= [value] \\ \int g(x)\,\mathrm{d}x &= [value] \end{aligned}\] The formula then gives \[\begin{aligned} \int_0^{[value]} [value] \,\mathrm{d}x &= \Bigg[\left([value]\right) \cdot \left([value]\right)\Bigg]_0^{[value]} - \int_0^{[value]} \left([value]\right) \cdot \left([value]\right) \,\mathrm{d}x\\ &= \Bigg[[value]\Bigg]_0^{[value]} - \Bigg[[value]\Bigg]_0^{[value]} \\ &= \left([value] - \left([value]\right)\right) - \left([value] - \left([value]\right)\right) \\ &= [value] \end{aligned}\]