# DEFINITE INTEGRAL

A definite integral is an integral in which limits are applied to the integral.

A definite integral is usually represented by the following

$$\int^{b}_{a}f(x) dx$$

Here, $a$ is known as the lower limit and $b$ is the upper limit.

For example,  If we are to calculate the increase in the value of the integral $3x^2$ from 1 to 3, it will be represented as:

$$\int^{3}_{1}3x^2 dx$$

The above can be calculated thus:

$\int^{3}_{1}3x^2 dx=\left[\frac{3x^{2+1}}{3}+C\right]^3_1$

$\left[x^3+C\right]^3_1$

Substituting,

$\left[x^3+C\right]^3_1=\left(3^3+C\right)-\left(1^3+C\right)$

$27+C-1-C=26$

Notice that the arbitrary constant, C, cancels out indefinite integral.

This tells you that the arbitrary Constant, C is not always added to definite integral because it will cancel out when the limit of the integral is applied.

Example 1

Evaluate $\int^4_2(x^2+4) dx$

Solution:

Using the power rule of integration

$\int^4_2(x^2+4)=\left[\frac{x^{2+1}}{3}+\frac{4x}{1}\right]^4_2$

$\left[\frac{x^{3}}{3}+4x\right]^4_2$

Substituting the upper and lower limit

$\left[\frac{x^3}{3}+4x\right]^4_2=\left(\frac{(4)^3}{3}+4(4)\right)-\left(\frac{(2)^3}{3}+4(2)\right)$

$\frac{64}{3}+16-\frac{8}{3}-8$

Taking L.C.D

$\frac{64-48-8-24}{3}=\frac{80}{3}=26$

Example 2

Given that $f(x)=3x^2-8x+4$. Find the $\int^3_2 f(x) dx$

Solution:

$\int^3_2 f(x) dx$ means we should find the definite integral of f(x) from 2 to 3

$\int^3_2 (3x^2-8x+4) dx=\left[\frac{3x^{2+1}}{3}-\frac{8x^{1+1}}{2}+\frac{4x^{0+1}}{1}\right]^3_2$.

$\left[x^3-4x^2+4x\right]^3_2$

By substitution

$\left(3^3-4(3)^2+4(3)\right)-\left(2^3-4(2)^2-4(2)\right)$

$\left(27-36+12\right)-\left(8-16+8\right)$

$3-0=3$

Example 3

Evaluate $\int^2_14e^{2x}$

Solution:

Recalled that $\int e^{ax}=\frac{e^{ax}}{a}$

$\int^2_14e^{2x}=\left[\frac{4e^{2x}}{2}\right]^2_1$

$\left[2e^{2x}\right]^2_1$

Applying the definite integral

$(2e^{2(2)})-(2e^{2(1))$

$2e^4-2e^2=109.2963-14.7781=94.42$

Example 4

Evaluate $\int^2_0 3\sin x dx$

Solution:

Since the derivative of $\cos x$ is $-\sin x$, it follows that the integral of $\sin x$ is $-\cos x$

$\left[-3 \cos x\right]^2_0$

$(-3 \cos 2)-(-3 \cos 0)$

$1.2484+3=4.2484$

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