A mineral deposit along a strip of length 7 cm has density s(x)=0.04x(7−x) g/cm for 0≤x≤7. Calculate the total mass of the deposit.
I guess you want
∫[0,7] s(x) dx = ∫[0,7] 0.04x(7-x) dx = 2.28667 g
To calculate the total mass of the mineral deposit, we need to integrate the density function over the given length.
The density function, s(x), is given by s(x) = 0.04x(7−x) g/cm.
To find the total mass, we integrate this density function over the interval from 0 to 7 cm.
The integral of the density function over the given interval is:
M = ∫[0,7] 0.04x(7−x) dx
To solve this integral, we can expand the expression inside the integral:
M = 0.04∫[0,7] (7x - x^2) dx
To integrate this expression, we can use the power rule of integration:
M = 0.04 [ (3.5x^2 - (1/3)x^3) ] evaluated from 0 to 7
Now, we substitute the upper and lower limits into the expression:
M = 0.04 [ (3.5(7)^2 - (1/3)(7)^3) - (3.5(0)^2 - (1/3)(0)^3) ]
M = 0.04 [ (3.5(49) - (1/3)(343)) - (0) ]
M = 0.04 [ (171.5 - 114.33) ]
M = 0.04 * 57.17
M = 2.2868 g
Therefore, the total mass of the mineral deposit is approximately 2.2868 grams.
To calculate the total mass of the deposit, we need to integrate the density function over the given length of the strip.
Let's first set up the integral. The total mass of the deposit can be calculated by integrating the density function, s(x), over the entire length of the strip from x=0 to x=7:
m = ∫[0,7] s(x) dx
In this case, the density function is given as s(x) = 0.04x(7-x) g/cm.
Therefore, the integral becomes:
m = ∫[0,7] 0.04x(7-x) dx
To evaluate this integral, we need to expand the expression, multiply it out, and then integrate it term by term. Let's simplify the expression:
m = 0.04 ∫[0,7] (7x - x^2) dx
Now, let's integrate each term separately:
m = 0.04 [ (7/2)x^2 - (1/3)x^3 ] evaluated from x=0 to x=7
Substituting the limits of integration:
m = 0.04 [ (7/2)(7)^2 - (1/3)(7)^3 ] - 0.04 [ (7/2)(0)^2 - (1/3)(0)^3 ]
Simplifying further:
m = 0.04 [ (7/2)(49) - (1/3)(343) ]
m = 0.04 [ 171.5 - 114.33 ]
m = 0.04 [ 57.17 ]
Finally, calculating the value:
m = 2.2868 grams
Therefore, the total mass of the deposit is approximately 2.2868 grams.