63–68. Definite integrals Evaluate the following definite inte...

Question

63–68. Definite integrals Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms.

∫₁/₈¹ dx/x√(1 + x²/³)

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says to evaluate the definite integral, and we're given the integral from 4 to 9 of DX divided by the quantity of X, multiplied by the square root of the quantity of 1 plus the square root of X. So we're asked to evaluate this integral, and we're given the integral from 4 to 9 of DX divided by the quantity of X multiplied by the square root of the quantity of 1, plus the square root of X in quantity. And so to evaluate this integral, we're gonna make a substitution. So we're gonna set U equal to the square root of X. That means that DU is going to be equal to 1 divided by the quantity of 2 multiplied by the square root of X and quantity DX. It's going to be more advantageous to make the substitutions to write X as a function of u. That means that X is going to be equal to U 2d and DX is going to be equal to 2UDU. So, making this substitution, we see that our integral is going to be equal to the integral from, and here we need to change our limits of integration. So when X is equal to 4, we have U is equal to the square root of 4, which is going to be 2. So that's gonna be our lower limit of integration. And for the upper limit of integration, when X is equal to 9, U is going to be equal to the square root of 9, which is 3. Then for the integram, we will have 2 UU divided by the quantity of U squared multiplied by the square root of the quantity of 1 plus u in quantity, and we can simplify this expression, so this is going to be equal to the integral from 2 to 3 of 2 divided by the quantity of u multiplied by the square root of the quantity of 1 plus u in quantity d u. The next we need to make another substitution. So, we're gonna set V equal to 1 plus U. That means that DV is going to be equal to DU, but this also means that U is going to be equal to V minus 1. So, making this substitution, we see that our integral is going to be equal to the integral from, and here we need to change our limits of integration again. So, when U is equal to 2 for the lower limit of integration, V is going to be equal to 1 + 2, which is going to give us 3. And for the upper limit of integration, when U is equal to 3, B is going to be equal to 1 + 3, which is 4. And then for the integrand, we will have 2 divided by the quantity of the quantity of the minus 1 in quantity multiplied by the square root of V in quantity DV. And we need to make one final substitution. And so we're gonna set T equal to the square root of V. That means that DT is going to be equal to 1 divided by the quantity of 2 multiplied by the square root of V and quantity DV. So making this substitution, We see that our integral is going to be equal to the integral, and here again, we need to change the limits in integration. So for the lower limit of integration, when V is equal to 3, T is going to be equal to the square root of 3, and for the upper limit of integration, when V is equal to 4, T is going to be equal to the square root of 4, which is 2. And then for the integram, we will have 4. Divided by the quantity of T2 minus 1 in quantity DT. Now, for the next step, we want to break this fraction up into two separate fractions using partial fractions. And notice that the denominator of a fraction is the difference of 2 squares, and so that means we can factor it. So we will have 4. Divided by the quantity of the quantity of t minus 1 in quantity multiplied by the quantity of T plus 1 in quantity, and we need to break this up into two separate fractions using partial fractions. So this is going to be equal to a divided by the quantity of T minus 1 in quantity, plus B divided by the quantity of T + 1 in quantity. So now we need to solve for A and B. So the next step is to multiply both sides of the equation by the quantity of T minus 1 and quantity multiplied by the quantity of T + 1 in quantity. In doing so, we will have that 4 is equal to a multiplied by the quantity of T + 1 in quantity, plus B multiplied by the quantity of T minus 1 in quantity. So, multiplying everything out on the right hand side, we'll see that 4 is equal to. A multiplied by T. Plus, a Plus B multiplied by T minus B. And so we want to do is look at the linear terms in both sides of the equation. So, for the linear terms, we see that A plus B has to be equal to 0, since we have no linear terms on the left-hand side of the equation. And if we look at the constant terms, we see that a minus B has to be equal to 4. So now we have two equations with two unknowns, and so we need to solve for A and B. And so if we add these two equations together, we'll eliminate the B variable. And so doing so, we will have that 2A is equal to 4. And divide both sides of this equation by 2, we'll get that A is equal to 2. And so now we can use result by substituting in for A into either of the R equations. And so we'll use the first equation, but we'll have 2 plus B has to be equal to 0, and so that means that B is going to be equal to minus 2. So, coming back to our interval. Our original integral, which was the integral from 4 to 9 of DX. Divided by the quantity of X multiplied by the square root of the quantity of 1 plus the square root of X in quantity is going to be equal to the integral from the square root of 3 to 2. Of 2 divided by the quantity of T minus 1 and quantity DT. Plus the integral from the square root of 3 to 2 of minus 2 divided by the quantity of T plus 1 in quantity DT. Now we just need to perform the integration. So this is going to be equal to. For the first integral, we're going to have 2 multiplied by, and here we're going to be integrating 1 divided by the quantity of T minus 1 in the quantity with respect to T, and that's going to give us the natural log of the absolute value of the quantity of T minus 1 in quantity. This needs to be evaluated from the square root of 3 to 2. And then we'll have minus 2 multiplied by, and here we're integrating 1 divided by the quantity of T + 1 in quantity with respect to T, and that's going to give us the natural log of the absolute value of the quantity of T + 1 in quantity, and this needs to be evaluated from the square root of 3 to 2. So substituting in our limits of integration, this is going to be equal to 2 multiplied by the quantity of the natural log of the absolute value of the quantity of 2 minus 1 in quantity, minus the natural log of the absolute value of the quantity of the square root of 3 minus 1. In quantity, then another in quantity, then we'll have minus 2. Multiplied by the quantity of the natural log of the absolute value of the quantity of 2 + 1 in quantity minus the natural log of the absolute value of the quantity of the square root of 3, plus 1 in quantity, then another in quantity. So simplifying this expression, this is going to be equal to. 2, multiplied by the quantity, and here for the first term, we'll have the natural log of, or sorry, natural log of 1, which is going to give us 0, so we'll have 0 minus. The natural log of the quantity, the square root of 3 minus 1 in quantity, then another in quantity, and here we drop the absolute value sign since the square root of 3 minus 1 is a positive quantity. Then we'll have minus 2 multiplied by the quantity of the natural log of 3 minus the natural log of the quantity of the square root of 3 plus 1 in quantity, then another in quantity. So, here again, we have dropped the absolute value sign since the arguments for both of the natural log functions are positive. So now what we want to do is to combine all of our natural log functions into a single natural log function. So doing so, this is going to be equal to 2 multiplied by the natural log of the quantity of the quantity of the square root of 3 plus 1 in quantity, divided by the quantity of 3 multiplied by the quantity of the square root of 3 minus 1 in quantity. To simplify this further, we're gonna multiply. The numerator and denominator of the argument of the natural log function by the square root 3 + 1. In doing so, this is going to be equal to 2 multiplied by the natural log of the quantity. Of 4 plus. 2 multiplied by the square root of 3, and then that quantity is going to be divided by the quantity of 6. In quantity And simplifying this further, we see that this is going to be equal to. 2 multiplied by the natural log of the quantity of 2 plus the square root of 3, and that quantity is going to be divided by 3 in quantity. And so this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

63–68. Definite integrals Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms.∫₁/₈¹ dx/x√(1 + x²/³)

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63–68. Definite integrals Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms.
∫₁/₈¹ dx/x√(1 + x²/³)