Definite Integrals

Definite Integrals

Properties: 

1. 

Example:

$$\int_2^5 x^2\,dx = -\int_5^2 x^2\,dx$$

LHS:$$= \left[ \frac{x^3}{3} \right]_2^5 = \frac{125 - 8}{3} = \frac{117}{3} = 39$$

RHS:

$$= -\left[ \frac{x^3}{3} \right]_5^2 = -39$$

2

  sums based on above properties👈

Example:    Evaluate

$$\int_{-3}^{3} |x|\,dx$$

Split at $x = 0$:

$$= \int_{-3}^0 (-x)\,dx + \int_0^3 x\,dx$$
$$= \left[ -\frac{x^2}{2} \right]_{-3}^0 + \left[ \frac{x^2}{2} \right]_0^3 = \left( 0 - \frac{9}{2} \right) + \left( \frac{9}{2} - 0 \right) = -\frac{9}{2} + \frac{9}{2} = 9$$3            Evaluate$$\int_{-2}^{3} |x^2 - 4|\,dx$$

We solve $x^2 - 4 = 0 \Rightarrow x = \pm2$

Break the integral into intervals where the expression inside modulus changes sign:

• 
  

$$\int_{-2}^{3} |x^2 - 4|\,dx = \int_{-2}^{2} -(x^2 - 4)\,dx + \int_{2}^{3} (x^2 - 4)\,dx$$

First part:

$$= \int_{-2}^{2} (4 - x^2)\,dx = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - (-8 + \frac{8}{3}) = 16 - \frac{16}{3} = \frac{32}{3}$$

Second part:

$$\int_2^3 (x^2 - 4)\,dx = \left[ \frac{x^3}{3} - 4x \right]_2^3 = \left( \frac{27}{3} - 12 \right) - \left( \frac{8}{3} - 8 \right) = (9 - 12) - \left( \frac{8}{3} - 8 \right) = -3 - \left( \frac{8}{3} - 8 \right) = -3 - \left( \frac{8 - 24}{3} \right) = -3 + \frac{16}{3} = \frac{7}{3}$$

Total:

$$\frac{32}{3} + \frac{7}{3} = \boxed{\frac{39}{3} = 13}$$

Symmetry: Even Function

If $f(-x)=f(x)\; then:$

$$\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$$

Example:

$$\int_{-2}^2 x^2\,dx = 2\int_0^2 x^2\,dx$$

LHS:

$$\left[ \frac{x^3}{3} \right]_{-2}^2 = \frac{8 - (-8)}{3} = \frac{16}{3}$$

RHS:

$$2 \cdot \left[ \frac{x^3}{3} \right]_0^2 = 2 \cdot \frac{8}{3} = \frac{16}{3}$$

Symmetry: Odd Function

If $\mathrm{<b>f</b>}<b>(</b><b>-</b>\mathrm{<b>x</b>}<b>)</b><b>=</b><b>-</b>\mathrm{<b>f</b>}<b>(</b>\mathrm{<b>x</b>}<b>) \; \; \; </b>\; \; \; then:$

$$\int_{-a}^a f(x)\,dx = 0$$

Example:

$$\int_{-3}^{3} x^3\,dx = 0$$6.   

Property:

$$\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$$

sums based on above properties👈

Example:

$$\int_0^1 \frac{x}{1 + x^2} \,dx = \int_0^1 \frac{1 - x}{1 + (1 - x)^2} \,dx$$7.  

property

sums based on above properties👈

Example: 

Now use the property:

$$\int_0^a f(x)\,dx + \int_0^a f(a - x)\,dx = \int_0^a [f(x) + f(a - x)]\,dx$$

Let’s find:

$$f(x) + f\left(\frac{\pi}{2} - x\right)$$

Let’s compute:

• $f(x)=\frac{\sin x}{\sin x+\cos \mathrm{x}}$

• $f\left(\frac{\pi}{2} - x\right) = \frac{\sin(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} = \frac{\cos x}{\cos x + \sin x}$

So:

$$f(x) + f\left(\frac{\pi}{2} - x\right) = \frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\sin x + \cos x} = \frac{\sin x + \cos x}{\sin x + \cos x} = 1$$

Thus:

$$f(x) + f\left(\frac{\pi}{2} - x\right) = 1$$

Now:

$$2I = \int_0^{\frac{\pi}{2}} \left[ f(x) + f\left(\frac{\pi}{2} - x \right) \right] dx = \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}$$

So:

$$I = \frac{\pi}{4}$$