Interesting that only integrals along the $x$ -axis are suggested when integrating along the $y$ -axis would be much simpler... Anyway, you have to split the interval of integration into two. The "height" of the region is not uniform over the entire interval.

When $y=0$ , we have $0^2=x-1\implies x=1$ . When $y=2$ , we have $2^2=x-1\implies x=5$ . Then the area we want is given by

$\displaystyle\int_0^12\,\mathrm dx+\int_1^52-\sqrt{x-1}\,\mathrm dx$

which seems to agree with the last option.

5 0

3 years ago

Find the following list of data calculate a demean be the maid and see mode or mothers for the following numbers listed in the p

Sholpan [36]

Answer:

Mean = 4.8875

Median = 4.6

Mode = 4.5 and 7.7

Step-by-step explanation:

Mean is the sum of total of data divided by the sample size

Sum total = 1.5 + 4.7 + 6 + 7.7 + 7.7 + 4.5 + 2.5 + 4.5

Sum total = 39.1

Sample size = 8

Mean = 39.1/8

Mean = 4.8875

To get the median we need to first rearrange

1.5, 2.5, 4.5, 4.5, 4.7, 6, 7.7, 7.7

Median = 4.5 + 4.7/2

Median = 4.6

Hence the median is 4.6

Mode is the value occuring the most. Since 4.5 and 7.7 both occurs twice, hence the mode of the data is 4.5 and 7.7

4 0

3 years ago

Convert the following sum into a product. cosx + cos3x + cos5x + cos7x

kvv77 [185]

Answer:

$4\cos x\cos (4x)\cos (2x)$

Step-by-step explanation:

Given: $cosx + cos3x + cos5x + cos7x$

To convert: the given sum into product

Solution:

Use formula: $\cos A+\cos B=2\cos \left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos (2x)\cos (-x)+2\cos (6x)\cos (-x)\\=2\cos (2x)\cos (x)+2\cos (6x)\cos (x)\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]\\=2\cos x\left [2 \cos \left ( \frac{2x+6x}{2} \right )\cos \left ( \frac{2x-6x}{2} \right ) \right ]\\=2\cos x\left [ 2\cos (4x) \cos (-2x) \right ]\\=4\cos x\cos (4x)\cos (2x)$