Definite Integrals

In Geometry, we know how to calculate the area of a triangular plot of land.

We can also find the area of the plot, if it is in the shape of a rectangle or a Square.

All the shapes mentioned above (triangle, rectangle, square) are regular shapes, whose area can be found out by the known results called formulas.

Now, if we are given a shape which is not regular or standard one, then how do we find its area? We try to approximate its area by associating it with regular shapes.

For finding the area of such irregular shapes, area enclosed between two geometrical shapes we make use of Definite Integrals.

Here, we introduce the concept of Definite Integrals.

A definite integral is denoted by $\int_{a}^{b}$ f(x)dx where a is called the lower limit of the integral & b is called the upper limit of the integral.
$\int_{a}^{b}$ f(x)dx denotes the area of the region bounded by the curve y = f(x), the ordinates x = a, x = band the x-axis.

Since the definite integral denotes the area enclosed, hence its value is always definite or fixed. So, it is called definite integral.
Relation between Indefinite & definite integral.

If f(x) be a continuous function defined on the closed interval [a, b] & F(x) is its antiderivative i.e. $\int_{}^{}$ f(x)dx = F(x) , then $\int_{a}^{b}$ f(x)dx = $[F(x)]_{a}^{b}$ = F(b) – F(a)

The above statement is called the Second fundamental Theorem of integral calculus.

Note: While evaluating a definite integral the Constant of Integration, C disappears at the end and hence its effect is nullified. So, we do not write the constant of integration while evaluating a definite integral.

Examples

Now let’s consider some examples on definite integrals .

Example 1: Evaluate the definite integral: $\int_{0}^{2}(x^2 +1)$ dx

Let F(x) = $\int$ $(x^2 + 1)$ dx = $\frac{x^{2+1}}{2+1}$ + x = $\frac{x^{3}}{3}$ +x

$\int_{0}^{2}(x^2 +1)$ dx = $[F(x)]_{0}^{2}$ = F(2) – F(0)

= $\left ( \frac{2^3}{3} +2 \right )$ – $\left ( \frac{0^3}{3} +0 \right )$ = $\left ( \frac{8}{3} +2 \right )$ – 0 = $\frac{14}{3}$

Note: The constant of Integration is not written here, as it does not change the final value of the definite integral.

Example 2: Evaluate: $\int_{-1}^{1}$ $(e^x +x)$ dx

Let F(x) = $\int$ $(e^x +x)$ dx = $e^x$ + $\frac{x^{1+1}}{1+1}$ = $e^x$ + $\frac{x^{2}}{2}$

$\int_{-1}^{1}$ $(e^x +x)$ dx = $[F(x)]_{-1}^{1}$ = F(1) – F(-1)

= $\left ( e^1+\frac{(1)^{2}}{2} \right )$ – $\left ( e^{-1}+\frac{(-1)^{2}}{2} \right )$ = e + $\frac{1}{2}$ – $e^{-1}$ – $\frac{1}{2}$

e – $e^{-1}$ = e – $\frac{1}{e}$

Example 3: Evaluate: $\int_{1}^{5}$ $x^2$ $\left ( 1 - \frac{1}{x^3} \right )$ dx

Let F(x) = $\int$ $x^2$ $\left ( 1 - \frac{1}{x^3} \right )$ dx = $\int$ $\left ( x^2 - \frac{1}{x} \right )$ dx

= $\int$ $x^2$ dx – $\int$ $\frac{1}{x}$ dx = $\frac{x^{2+1}}{2+1}$ – log |x| = $\frac{x^3}{3}$ – log|x|

$\int_{1}^{5}$ $x^2$ $\left ( 1 - \frac{1}{x^3} \right )$ dx = $[F(x)]_{1}^{5}$ = F(5) – F(1)

= $\left ( \frac{5^3}{3}-log|5| \right )$ – $\left ( \frac{1^3}{3}-log|1| \right )$ = $\left ( \frac{125}{3}-log|5| \right )$ – $\left ( \frac{1}{3}-log|1| \right )$

= $\left ( \frac{125}{3}-\frac{1}{3} \right )$ – log|5| + log|1| = $\frac{124}{3}$ – log|5| + 0 = $\frac{124}{3}$ – log|5|

Example 4: Evaluate: $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}$ $\frac{3+4Cosx}{Sin^2x}$ dx

Let F(x) = $\int$ $\frac{3+4Cosx}{Sin^2x}$ dx = $\int$ $\left ( \frac{3}{Sin^2x} + 4 \frac{Cosx}{Sin^2x} \right )$ dx

= $\int$ $(3Cosec^2x + 4CosecxCotx)$ dx = 3 $\int$ $Cosec^2x$ dx + 4 $\int$ CosecxCotx

= -3Cotx – 4Cosecx

$\int_{0}^{\frac{\pi}{2}}$ $\frac{3+4Cosx}{Sin^2x}$ dx = $[F(x)]_{0}^{\frac{\pi}{2}}$ = F $\left ( \frac{\pi}{2} \right )$ – F $\left ( \frac{\pi}{4} \right )$

$\left ( -3Cot\frac{\pi}{2} - 4Cosec\frac{\pi}{2} \right )$ – $\left ( -3Cot\frac{\pi}{4} - 4Cosec\frac{\pi}{4} \right )$

= (-3(0)-4(1)) – $\left ( -3(1)-4(\sqr2) \right )$

= (0 – 4) – $\left ( -3-4\sqr{2} \right )$ = 4 + 3 +4 $\sqr{2}$ = -1 +4 $\sqr{2}$

Example 5: Evaluate $\int_{1}^{4}$ $\left ( 3x^2 -2x +7\sqr{x} \right )$ dx

Let F(x) = $\int$ $\left ( 3x^2 -2x +7\sqr{x} \right )$ dx

= 3 $\int$ $x^2$ dx – 2 $\int$ xdx + 7 $\int$ $x^{1/2}$ dx = 3 $\frac{x^{2+1}}{2+1}$ – 2 $\left ( \frac{x^{1+1}}{2} \right )$ + 7 $\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}$ +C

= $x^3$ – $x^2$ + $\frac{14}{3}$ $x^{\frac{3}{2}}$ = $x^3$ – $x^2$ + $\frac{14}{3}$ x $\sqr{x}$

$\int_{1}^{4}$ $\left ( 3x^2-2x+7\sqr{x} \right )$ dx = $[F(x)]_{1}^{4}$ = F(4) – F(1)

= $\left ( 4^3 - 4^2 +\frac{14}{3}4\sqr{4} \right )$ – $\left ( 1^3 - 1^2 +\frac{14}{3}(1)\sqr{1} \right )$

= $\left ( 64 - 16 +\frac{112}{3}\right )$ – $\left ( 0 +\frac{14}{3}\right )$ = 48 + 98 = $\frac{144-98}{3}$ = $\frac{46}{3}$

Check Point

Evaluate the following definite integrals:

$\int_{0}^{1}$ (x + 1)dx
$\int_{0}^{\frac{\pi}{4}}$ $\left ( 7Cosx - 3Sec^2x \right )$ dx
$\int_{0}^{1}$ $\left ( 7e^x + 34x^{16}+17 \right )$ dx
$\int_{1}^{2}$ (x – 1)(x – 2)dx
$\int_{1}^{4}$ $\left ( \sqr{x} - \frac{1}{\sqr{x}} \right )$ dx

Answer Key

$\int_{0}^{1}$ (x + 1)dx = $\frac{3}{2}$
$\int_{0}^{\frac{\pi}{4}}$ $\left ( 7Cosx - 3Sec^2x \right )$ dx =
$\int_{0}^{1}$ $\left ( 7e^x + 34x^{16}+17 \right )$ dx = 7e +19
$\int_{1}^{2}$ (x – 1)(x – 2)dx = $-\frac{1}{6}$
$\int_{1}^{4}$ $\left ( \sqr{x} - \frac{1}{\sqr{x}} \right )$ dx = $\frac{17}{2}$

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