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Definite Integral by Parts is a special case of Definite Integral. The LIATE Rule plays an important role in solving integration by parts in definite integration. In this article, we will learn about Definite Integral by Parts, How to Solve with LIATE Rule, Solved Examples, Properties, and Applications of Definite Integrals by Parts.

Integration is a method of uniting the parts together to form a whole. We find a function whose differential is known and we proceed to find its integrals.

The method of determining integrals is termed integration. By parts, definite integrals are applied where the limits are defined and indefinite integrals are executed when the boundaries of the integrand are not defined. Definite Integration by Parts is similar to integration by parts of indefinite integrals. Definite integration by parts is used when the function is a product of two terms of the independent variable. One term is called as u and another term is called as v. The u and v terms are decided by LIATE rule.

The function that we are supposed to integrate must be continuous between the range, that is there should not be any gaps, drops or vertical asymptotes where the function goes up/down towards infinity.

The notation of the Integral is as shown below:

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The following steps are used in Definite Integration by Parts

The LIATE Rule is as follows

A rule of thumb has been proposed, consisting of choosing as u the function that comes first in the following list:

The function which is to be v is whichever comes last in the list. The reason is that functions lower on the list generally have easier antiderivatives than the functions above them.

\(\int_{a}^{ b}uvdx=u\int_a^b{v}dx-\int_{a}^{b}u’vdx\)

All we have to do is evaluate the term uv for limit b(upper limit) and subtract off the evaluation for limit a(lower limit).

Similarly, perform the remaining half and end the process by substitution of limits and taking the difference of the values at the upper and lower limits.

All the properties of Definite Integral are applicable for Definite Integral by Parts.

Learn about Double Integral

The applications of Definite Integrals by Parts are as follows

Now let’s see some solved examples on definite integration by parts.

Solved Example 1: I = \(\int_{-1}^{2}x.e^{6x}dx\)

Solution:

I = \(\int_{-1}^{2}x.e^{6x}dx\)

u=x and \(v=e^{6x}\)

a = -1 and b = 2

Therefore, by using \(\int_{a}^{ b}uvdx=u\int_a^b{v}dx-\int_{a}^{b}u’vdx\)

We get,

I = \({[{xe^{6x}\over{6}}]_{-1}^2}-\int_{-1}^{2}1.{e^{6x}\over{6}}dx\)

I = \({11\over{36}}e^{12}-{1\over{36}}[e^{6x}]_{-1}^2\)

I = \({11\over36}e^{12}-{7\over36}e^{-6}\)

Solved Example 2: I = \(\int_0^\pi(3t+5)cos({t\over{4}})dt\)

Solution: I = \(\int_0^\pi(3t+5)cos({t\over{4}})dt\)

u=(3t+5) and \(v=cos({t\over{4}})\)

a = 0 and b = \(\pi\)

Therefore, by using \(\int_{a}^{ b}uvdx=u\int_a^b{v}dx-\int_{a}^{b}u’vdx\)

We get,

\(I = [(3t+5)(-sin({t\over{4}}))]_0^\pi.{1\over{1\over4}}-\int_{0}^{\pi}3.(-sin({t\over{4}})).{1\over{1\over4}}dx\)

\(I = -(3\pi+5)2\sqrt{2}+12{(cos({t\over{4}}))}.{1\over{1\over4}}]^\pi_0\)

\(I = -(3\pi+5)2\sqrt{2}+48[{1\over{\sqrt{2}}}-1]\)

I = \(\begin{matrix}{l}\int_1^ex\log x\ dx\\=\left\{\log x.\ \frac{x^2}{2}\right\}_1^e\ -\int_1^e\frac{1}{x}.\ \ \frac{x^2}{2}\ dx\\=\left\{\frac{x^2\log x}{2}\right\}_1^e\ -\frac{1}{2}\int_1^e\ x\ dx\\=\left\{\frac{e^2\log e}{2}-\frac{\log1}{2}\right\}\ -\left(\ \frac{x^2}{4}\right)_1^e\\=\left\{\frac{e^2}{2}\right\}\ -\left\{\frac{e^2}{4}-\frac{1}{4}\right\}\\=\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4}\\=\frac{e^2+1}{4}=2.097\end{matrix}\)

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