Definite integration by parts is a special method used to solve definite integrals where two functions are multiplied together. To decide which function to choose first, we use a helpful trick called the LIATE rule. This rule helps us pick the right order for easier solving. In this method, we break one function down and integrate the other. Definite integration by parts helps find the exact value of an integral between two limits. It’s useful in many real-world problems. In simple terms, integration helps us combine small parts to get the total or whole value of a quantity.

Definite Integration by Parts

Integration is the process of finding the total or overall value from small parts. When we use definite integration by parts, we solve integrals that have specific starting and ending points, called limits. This method is helpful when the function we’re working with is a product of two different expressions involving the variable. These two parts are named u and v, and we choose them using the LIATE rule, which gives a guideline for which function to pick first.

Definite integration by parts works similarly to the regular (indefinite) integration by parts, but here we apply limits at the end to get a final value. It's important that the function we want to integrate is continuous between the given limits. That means the graph of the function should not have breaks, jumps, or go off to infinity within that range. This ensures the result of the integration is accurate and meaningful.

The notation of the Integral is as shown below:

The following steps are used in Definite Integration by Parts

• Choose u and v by LIATE rule explained below
• Find the Differential of u: u’
• Find the Integral of v: ∫v dx.
• Put u, u’ and ∫v dx into: u∫v dx −∫u’ (∫v dx) dx.
• Simplify and solve.

LIATE Rule
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:

L – logarithmic functions: ln(x), logb(x), etc.
I – inverse trigonometric functions (including hyperbolic analogues): arctan(x), arcsec(x), arsinh(x), etc.
A – algebraic functions: x², 3x⁵⁰, etc.
T – trigonometric functions (including hyperbolic analogues): sin(x), tan(x), sech(x), etc.
E – exponential functions: eˣ, 19ˣ, etc.

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.

∫ from a to b of u·v dx = u·∫ from a to b of v dx − ∫ from a to b of u′·v dx

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.

Properties of Definite Integrals by Parts

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

The value of a definite integral does not vary with the change of the variable of integration when the limits of integration remain the same.
∫[a to b] f(x) dx = ∫[a to b] f(t) dt

When the limits of integration are changed, the sign of integral also changes.
∫[a to b] f(x) dx = −∫[b to a] f(x) dx

∫[a to a] f(x) dx = 0

∫[a to b] f(x) dx = ∫[a to c] f(x) dx + ∫[c to b] f(x) dx, where a < c < b

∫[a to b] f(x) dx = ∫[a to b] f(a + b − x) dx

∫[0 to a] f(x) dx = ∫[0 to a] f(a − x) dx

∫[0 to 2a] f(x) dx = ∫[0 to a] f(x) dx + ∫[0 to a] f(2a − x) dx, if f(2a − x) = f(x)

∫[0 to 2a] f(x) dx = 2∫[0 to a] f(x) dx, if f(2a − x) = f(x)
and
∫[0 to 2a] f(x) dx = 0, if f(2a − x) = −f(x)

∫[−a to a] f(x) dx = 2∫[0 to a] f(x) dx, if f is an even function i.e. f(−x) = f(x)
∫[−a to a] f(x) dx = 0, if f is an odd function i.e. f(−x) = −f(x)

Learn about Double Integral

Applications of Definite Integrals by Parts

The applications of Definite Integrals by Parts are as follows

• Integration by parts is often used in harmonic analysis, especially in Fourier analysis. It is used to represent those quickly oscillating integrals with sufficiently smooth integrands decay quickly. The most common example of this is its use in showing that the decay of function’s Fourier transform depends on the smoothness of that function.
• One use of integration by parts in operator theory is that it shows that the −∆ (where ∆ is the Laplace operator) is a positive operator on L. If f is smooth and compactly supported then we use integration by parts.
•  Definite Integrals by Parts is used for deriving the Euler–Lagrange equation in the calculus of variations.

Solved Examples of Definite Integral by Parts

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

Solved Example 1:

I = ∫[−1 to 2] x·e^(6x) dx

Solution:
I = ∫[−1 to 2] x·e^(6x) dx
Let:
u = x ⇒ du = 1
dv = e^(6x) dx ⇒ v = e^(6x)/6

Using the integration by parts formula:
∫[a to b] u·v dx = [u·v] from a to b − ∫[a to b] v·du

We get:
I = [x·e^(6x)/6] from −1 to 2 − ∫[−1 to 2] (1·e^(6x)/6) dx
I = [2·e^(12)/6 − (−1)·e^(−6)/6] − (1/6)∫[−1 to 2] e^(6x) dx
I = (2e^12 + e^−6)/6 − (1/6)·[e^(6x)/6] from −1 to 2
I = (2e^12 + e^−6)/6 − (1/36)(e^12 − e^−6)
I = (12e^12 + 6e^−6 − e^12 + e^−6)/36
I = (11e^12 + 7e^−6)/36

Solved Example 2:

I = ∫[0 to 1] x·e^(2x) dx

Solution:

We are given:
I = ∫[0 to 1] x·e^(2x) dx

Let:
u = x ⇒ du = 1
dv = e^(2x) dx ⇒ v = e^(2x)/2

Using the integration by parts formula:
∫[a to b] u·v dx = [u·v] from a to b − ∫[a to b] v·du

Now apply the values:
I = [x·e^(2x)/2] from 0 to 1 − ∫[0 to 1] (1·e^(2x)/2) dx
I = [1·e^2/2 − 0·e^0/2] − (1/2)∫[0 to 1] e^(2x) dx
I = (e^2/2) − (1/2)·[e^(2x)/2] from 0 to 1
I = (e^2/2) − (1/4)(e^2 − 1)
I = (e^2/2) − (e^2/4 − 1/4)
I = e^2/4 + 1/4
I = (e^2 + 1)/4

Solved Example 3:
I = ∫[1 to e] x·log(x) dx
= {log(x)·x²/2} from 1 to e − ∫[1 to e] 1/x · x²/2 dx
= {x²·log(x)/2} from 1 to e − (1/2)∫[1 to e] x dx
= {e²·log(e)/2 − log(1)/2} − (x²/4) from 1 to e
= e²/2 − (e²/4 − 1/4)
= e²/2 − e²/4 + 1/4
= e²/4 + 1/4
= (e² + 1)/4
≈ 2.097

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