A mineral deposit along a strip of length 8 cm has density s(x)=0.07x(8−x) g/cm for 0≤x≤8. Calculate the total mass of the deposit. Your answer must include units.
.07 [ 8 x dx - x^2 dx ] at 8 - at 0
.07 [ 8 x^2/2 - x^3/3 ]
.07 [ 4*64 - 512/3 ]
(.07/3) [12*64 -512] = (.07/3)[256] grams
Why did the mineral deposit bring a calculator to the strip of length 8 cm?
Because it wanted to calculate its total mass, of course! Let's do the math.
To find the total mass of the deposit, we need to calculate the integral of the density function over the interval 0≤x≤8.
So, let's integrate the density function s(x) = 0.07x(8−x) g/cm over the interval 0≤x≤8.
∫[0,8] 0.07x(8−x) dx
Using the properties of integrals, we can expand and solve it step by step:
∫[0,8] (0.07x * 8 - 0.07x^2) dx
= 0.07 * ∫[0,8] (8x - x^2) dx
= 0.07 * (4x^2 - (x^3/3)) | [0,8]
Now we substitute the limits of integration:
= 0.07 * [(4(8)^2 - ((8)^3/3)) - (4(0)^2 - ((0)^3/3))]
Calculating further:
= 0.07 * [(4 * 64 - (512/3)) - (0 - 0)]
= 0.07 * [ (256 - (512/3))]
= 0.07 * (256 - 170.67)
= 0.07 * 85.33
= 5.97 g
Therefore, the total mass of the deposit is 5.97 grams. The mineral deposit is pretty lightweight for a strip of that length!
To calculate the total mass of the deposit, we need to integrate the density function over the given length (0 to 8 cm).
The density function is given as s(x) = 0.07x(8 - x) g/cm.
We can express the total mass (M) of the deposit as the integral of the density function:
M = ∫ s(x) dx
Now, let's calculate the integral:
M = ∫ 0.07x(8 - x) dx
To solve this integral, we can expand the expression and then integrate:
M = 0.07 ∫ (8x - x^2) dx
M = 0.07 [ (8/2)x^2 - (1/3)x^3 ] from 0 to 8
M = 0.07 [ (4)x^2 - (1/3)x^3 ] from 0 to 8
M = 0.07 [ (4)(8)^2 - (1/3)(8)^3 ] - 0.07 [ (4)(0)^2 - (1/3)(0)^3 ]
M = 0.07 ( 256 - (1/3)(512) )
M = 0.07 ( 256 - 170.67 )
M = 0.07 ( 85.33 )
M ≈ 5.9731 g
Therefore, the total mass of the deposit is approximately 5.9731 grams.
To calculate the total mass of the mineral deposit, we need to integrate the density function over the given interval.
The density function is given by s(x) = 0.07x(8−x) g/cm for 0≤x≤8.
We can integrate the density function s(x) with respect to x from 0 to 8 to find the total mass. The integral is given by:
∫[from 0 to 8] 0.07x(8−x) dx
To find the integral, we need to expand the expression:
∫[from 0 to 8] (0.07x)(8) - (0.07x)(x) dx
Next, we can distribute 0.07x to both terms:
∫[from 0 to 8] 0.56x - 0.07x^2 dx
Now, let's integrate each term separately:
∫[from 0 to 8] 0.56x dx - ∫[from 0 to 8] 0.07x^2 dx
The integral of 0.56x with respect to x is 0.28x^2, and the integral of 0.07x^2 with respect to x is 0.07(1/3)x^3 = 0.023333x^3.
Evaluating the integrals from 0 to 8, we get:
[0.28x^2] evaluated from 0 to 8 - [0.023333x^3] evaluated from 0 to 8
Substituting the values:
[0.28(8^2) - 0.28(0^2)] - [0.023333(8^3) - 0.023333(0^3)]
Simplifying the calculations:
[0.28(64) - 0.28(0)] - [0.023333(512) - 0.023333(0)]
= [17.92 - 0] - [11.893333 - 0]
= 17.92 - 11.893333
= 6.026667
Therefore, the total mass of the mineral deposit is 6.026667 grams (g).