Find the area of the shaded region between f ( x ) = 1 x f\left(x\right)=\frac{1}{x} & g ( x ) = x g\left(x\right)=x from x = 0.5 x=0.5 to x = 4 x=4 .

this problem, we're asked to find the area of the shaded region between f of x equals one over x and g of x equals x, specifically on the interval from x equals 0.5 to x equals four. Now taking a look at our graph here, we can see that this is going to take two integrals in order to find this area because our function f of x starts out on top but then becomes the bottom function as it crosses paths with g of x. So we're gonna go ahead and label those interval points 0.54 on our graph here. This is 0.5, and then up here is x equals four. So we need to set up an integral that goes from 0.5 to one and then from one to four in order to find this area. So starting out here, our area is going to be equal to the integral from 0.5 to one, that's that first little sliver of area there, of our top function, which in this case is f of x, which is one over x. So one over x minus our bottom function, in this case, g of x, which is just x. So one over x minus x d x. That's that first integral. Then we're adding this together with this larger chunk of area here that is bounded from one to four of what is now our top function x minus our bottom function one over x and then d x. So now we just have to evaluate these integrals. So let's go ahead and get started there. Now we know that the antiderivative of one over x is the natural log of x. So this is the natural log of x. And then using the power rule here, this is minus one half x squared bounded from 0.5 to one. Then adding this together, the integral from one to four of x minus one over x, this is one half x squared minus the natural log of x bounded from one to four. So now we just need to plug those bounds in in order to ultimately find this area. So starting with this first part of our answer here, this is going to be the natural log of one minus one half times one squared. So that's just that upper bound of that first integral plugged in, and then we're subtracting off that lower bound plugged in. So this is the natural log of 0.5 minus one half times 0.5 squared. So this represents just that first integral with those balance plugged in. Now for our second integral here, I'm adding this together with a one half times four squared minus the natural log of four. That's that upper bound. And then minus one half times one squared minus the natural log of one. So this represents that second integral with those bounds plugged in. So now with those bounds plugged in, we just need to go ahead and clean this up. If at this point you just want to type this all into your calculator in order to get your final answer as a decimal, you can do that. Just go ahead and be careful with where you're putting parentheses and make sure you're using enough parentheses. But I'm gonna go ahead and keep going through this algebraically here. Now the natural log of one is zero. So both instances of that in our integrals can cancel out to zero here and then zero minus one half one squared and then minus all of this stuff over here. Everything that's left over in this part of our integral is going to be worked out to be negative three over eight minus the natural log of 0.5. So if we give everything a common denominator here, combine all of our like terms into one fraction and then have that natural log, this is everything that's left over. So that's everything in that yellow highlighter simplified down. Now doing the same thing with everything that is highlighted in green up here, if I go ahead and combine all my like terms, give everything a common denominator, we're going to get here 15 over two minus the natural log of four. So this is everything in green simplified down. Now, doing our final algebraic steps here, we need to combine both of these fractions, negative three eights and 15 over two. Negative threeeight and 15 over two. 15 over two with a common denominator of eight here is 60. 60 minuteus three is 57. So those two fractions together combine to 57 over eight. Then if I take the natural log of 0.5 and the natural log of four, since we have both of them as negative we can factor that negative out so it becomes negative natural log of 0.5 plus the natural log of four. If I have both of those in parentheses there being multiplied, that means that I can just multiply what those arguments are. 0.5 times four is two, so this becomes minus the natural log of two just using our log rules. So this is our final answer if we are not giving it as a decimal, if we just want it in its fully simplified algebraic form. But from here, if you type this into your calculator, get a decimal as an answer, you should get here 6.432 and this represents our shaded area in between our two functions on the interval from point five to four. If you have any questions feel free to let us know in the comments and I'll see you in the next video.

9. Graphical Applications of Integrals

Area Between Curves

Business Calculus

9. Graphical Applications of Integrals

Area Between Curves

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Multiple Choice

Find the area of the shaded region between $f (x) = \frac{1}{x}$ & $g (x) = x$ from $x equals 0.5$ to $x equals 4$ .

-5.796

6.432

0.329

0.557

Verified step by step guidance

Step 1: Understand the problem. We are tasked with finding the area of the shaded region between two functions, f(x) = 1/x and g(x) = x, over the interval [0.5, 4]. The shaded region is bounded by these two curves.

Step 2: Set up the integral. The area between two curves is calculated using the formula: A = ∫[a, b] (upper function - lower function) dx. Here, f(x) = 1/x is the upper function and g(x) = x is the lower function over the interval [0.5, 4].

Step 3: Write the integral expression. The area can be expressed as: A = ∫[0.5, 4] (1/x - x) dx. This represents the difference between the two functions integrated over the given interval.

Step 4: Break down the integral. Split the integral into two parts: A = ∫[0.5, 4] (1/x) dx - ∫[0.5, 4] (x) dx. This allows us to compute each term separately.

Step 5: Solve each integral. For ∫(1/x) dx, the antiderivative is ln|x|. For ∫(x) dx, the antiderivative is (x^2)/2. Substitute the limits of integration [0.5, 4] into each antiderivative to compute the area.