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3 years ago

What does this integral unit mean?

Mathematics

2 answers:

schepotkina [342]3 years ago

8 0

Answer:its an integral where the function to be integrated is evaluated along a curve. Various different line integrals are in use. In the case of a closed curve it is also called a contour integral. The function to be integrated may be a scalar field or a vector field.

Step-by-step explanation:

hope this helps

stepan [7]3 years ago

7 0

Answer:

Every time a new piece of equipment is added to the system, if ifs not properly optimized within the scope of the entire system, you'll end up with wasted energy and operational inefficiencies

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M is inversely proportional to r if m= 9 when r= 4 find m when r=2

bazaltina [42]

m=18 when r = 2.

Step-by-step explanation:

Given,

m∝ $\frac{1}{r}$

So,

m = k× $\frac{1}{r}$ ,--------eq 1, here k is the constant.

To find the value of m when r = 2

At first we need to find the value of k

Solution

Now,

Putting the values of m=9 and r = 4 in eq 1 we get,

9 = $\frac{k}{4}$

or, k = 36

So, eq 1 can be written as m= $\frac{36}{r}$

Now, we put r =2

m = $\frac{36}{2}$

or, m= 18

Hence,

m=18 when r = 2.

4 0

3 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$