Since $f(x)=3x+1$ is a linear function, its graph is a line. This means that you can decompose the area under its graph as the sum of a right triangle and a rectangle, as shown in the attached image.

All the dimensions are fairly easy to deduce: we have

$AC = 2,\ CB=6$

So, the area of the triangle is

$A_t = \dfrac{2\times 6}{2}=6$

For the rectangle, we have

$A_r = AC\times CE = 2\times 4=8$

So, the total area is $6+8=14$

Second integral

We have

$y = \sqrt{1-x^2} \implies y^2=1-x^2 \implies x^2+y^2=1$

So, this function is the upper half of the unit circle (the two halves are the functions $y=\pm\sqrt{1-x^2}$ )

Since the area of the unit circle is $\pi$ , the value of the integral is $\frac{\pi}{2}$

Third integral

It follows the exact same logic as the first one, you only have to adjust the numbers.

4 0

3 years ago

Benford's Law states that the probability that the first decimal-digit of a raw data sample (from 1 to 9) is given

patriot [66]

Answer: P(4) = 0.097

$\bold{\sum_{m=1}^9=1}$

<u>Step-by-step explanation:</u>

$P(m)=\log(m+1)-\log(m)\\\\.\qquad =\log\bigg(\dfrac{m+1}{m}\bigg)\qquad \text{(using rules for condensing logs)}\\\\\\P(4)=\log\bigg(\dfrac{4+1}{4}\bigg)\\\\\\.\qquad =\log\bigg(\dfrac{5}{4}\bigg)\\\\\\.\qquad =\large\boxed{0.097}$

$\sum_{m=1}^9\log\bigg(\dfrac{m+1}{m}\bigg)\\\\\\.\qquad =\log\bigg(\dfrac{2}{1}\bigg)+\log\bigg(\dfrac{3}{2}\bigg)+...\log\bigg(\dfrac{9}{8}\bigg)+\log\bigg(\dfrac{10}{9}\bigg)\\\\\\.\qquad =\log\bigg(\dfrac{2}{1}\bigg)\bigg(\dfrac{3}{2}\bigg)...\bigg(\dfrac{9}{8}\bigg)\bigg(\dfrac{10}{9}\bigg)\qquad \text{(using rules for condensing logs)}\\\\\\.\qquad =\log\bigg(\dfrac{10!}{9!}\bigg)\\\\\\.\qquad =\log(10)\\\\\\.\qquad =\large\boxed{1}$