Calculus integral related questions

Question

Let f be the function defined by f(x) = −ln(x) for 0 < x ≤ 1. R is the region between the graph of f and the x-axis.

1. This question does not permit the use of a calculator. Determine whether region R has a finite area. If so, find the area. If not, explain why.
2. This question does not permit the use of a calculator. Determine whether the solid generated by revolving region R about the y-axis has a finite volume. If so, find the volume. If not, explain why.
3. The formula for the arc length from a to b of a differentiable function is . Find the length of the curve f(x) from x = 0.5 to x = 1. Your work must show the integral, but you may use your calculator to evaluate it. Give 3 decimal places in your answer.
4. g(x) = x². Find the area bounded by f(x), g(x) and the x-axis in the interval [0.5, 1]. Your work must show the integral, but you may use your calculator to evaluate it. Give 3 decimal places in your final answer.
5. Find the volume of the solid whose base is the region bounded by f(x), g(x) and the x-axis on the interval [0, 1], and whose cross-sections perpendicular to the y-axis are squares. Your work must show the integral, but you may use your calculator to evaluate it. Give 3 decimal places for your answer.

Answer 1

Given that f(x) = -ln(x) for 0 < x ≤ 1, we know that ln(0) = -∞, so -ln(0) = ∞, and ln(1) = 0

The area of this function with respect the x-axis can be written as

int_{lb=0}_{ub}=1 (-ln(x))dx

But, the lower bound as we noted above is undefined (∞). So we cannot use zero as the lower bound, but can use any other number (however close to 0, but not zero) that we wish in order to have a finite area...

int_{lb=ε}_{ub}=1 (-ln(x))dx where ε > 0

So since x much be strictly > , the region would have a finite area

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Revolving this region about the y-axis is akin to calculating the annuli of the function (cylinders with holes in the middle) sometimes called the "cylindrical shell method". Area = 2*π*r*h. We just need to figure out r and h

Volume = int_{lb=ε}_{ub=1} (2*π*(x)*(-ln(x))dx

This can be done via integration by parts and yields...

Volume = -2*π* (x²ln(x)/2 - x²/4) evaluated at the bounds...

Volume = -2*π* ( [(1²*0 )/ 2 - 1²/4] - [ε²ln(ε)/2 - ε²/4] )

Volume = -2*π* (-1/4 + ε²/4 - ε²ln(ε))

Volume = 2*π*( (1-ε²)/4 + ε²ln(ε) )

In the limit as epsilon goes to zero (lim ε --> 0), the volume becomes π/2

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Arc Length as shown in the problem relates to the derivate of the function...

f(x) = -ln(x)

f'(x) = -1/x

L = int_{lb=0.5}_{ub=1} (sqrt(1 + 1/x²))dx

L = int_{lb=0.5}_{ub=1} ( sqrt(x²/(1 + x²)) ) dx = int_{lb=0.5}_{ub=1} ( x * sqrt(1/(1 + x²)) ) dx

L = sqrt(1 + x²) from 0.5 to 1

L = sqrt(1 + 1²) - sqrt(1+ 0.5²) = sqrt(2) - sqrt(5)/2

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Now we need to know where the intersection of the g(x) and f(x) resides for our integral to compute the area between the curves. Solving the transcendental equation (use desmos to plot both f(x) and g(x)) gives the intersection of these curves at approximately 0.653 in x. Let's call this R

Therefore, we use the additivity law for integrals...

int_{lb=0.5}_{ub=1} = int_{lb=0.5}_{ub=R} + int_{lb=R}_{ub=1}

Since g(x) < f(x) for x < R and g(x) > f(x) for x > R we have...

int_{lb=0.5}_{ub=R} (f(x) - g(x))dx + int_{lb=R}_{ub=1} (g(x) - f(x))dx

The first integral --> int_{lb=0.5}_{ub=R} (-ln(x) - x^2)dx = -xln(x) - x - x³/3 eval from 0.5 to R

= -Rln(R) - R - R³/3 - ln(2)/2 + 0.5 + (0.5)³/3

= -Rln(R) - R - R³/3 + (1 - ln(2))/2 + 1/24

The second integral --> int_{lb=R}_{ub=1} (g(x) - f(x))dx = x³/3 + xlnx + x eval from R to 1

= 1/3 + 1* 0 + 1 - R³/3 - Rln(R) - R

= 4/3 - R³/3 - Rln(R) - R

Adding them together yields --> 4/3 + 1/24 + (1-ln(2))/2 - 2*(R³/3 + Rln(R) + R)

Now plug in the value of R from above and approximate your response to 3 decimal places

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