Limit definition of the definite integral Use...

Question

Limit definition of the definite integral Use the limit definition of the definite integral with right Riemann sums and a regular partition to evaluate the following definite integrals. Use the Fundamental Theorem of Calculus to check your answer.

∫₀⁴ (𝓍³―𝓍) d𝓍

∫₀⁴ (𝓍³―𝓍) d𝓍

Accepted Answer

Hello. In this video, we are going to use the limit definition of the definite integral with the right Riemann sums to evaluate the definite integral from 0 to 5 of 2 X cubed minus 5 XDX. Now, the definition is given as the following. The integral from A to B of F of XDX is equal to the limit as N approaches infinity of the summation. Of Is equal to 1 ton of F of Xi multiplied by delta X. This is where Delta X is defined as B minus A divided by N and Xi is defined as A plus I delta X. So, what we are going to do is we are first going to identify our A and B values by looking at the given definite integral. Now, the definite integral is the integral from 0 to 5. A represents the lower boundary, which is going to be 0, and B represents the upper boundary, which is going to be 5. Now, in this case here, we can use these two values to identify our delta X value for the summation. Delta X is B minus A divided by N, which is 5 minus 0 divided by N, which further simplifies to be 5 divided by N. Now what we need to do is we need to solve the Xi term to plug into the given function. Xi is defined as A + I delta X, and that is going to be 0 plus the index I multiplied by delta X, which is 5 divided by N. And this is going to simplify to be 5 I divided by N. So this is going to be the value that we plug into every X for the given function in the integral. And now that we have the two components that we need, we can construct the right Riman integral. So this is going to be the integral. As the limit as N approaches infinity of I is equal to 1 ton of 2 multiplied by Xi, which is 5 I divided by n cubed, minus 5 multiplied by 5 I divided by N, and this whole quantity is going to be multiplied by delta X, which is 5 divided by N. Now what we are going to do is we are going to simplify the inside of the summation. By taking the product of all the terms together, we get the summation. From 1 ton. Of 250 I cubed divided by n cubed minus 25 I divided by N, and this is going to be multiplied by 5 divided by N. And by distributing 5 divided by N, we can further rewrite the summation to be the summation from I is equal to 1 at N of 1250 I cubed divided by N4 minus 125 I divided by n quad. This is going to be the right Ryman representation for the given definite integral. And now what we are going to do is we are going to evaluate this summation. So, in the first summation, the only thing we want to leave within the summation itself is going to be the I terms. So we are going to distribute the summation and move all the non-I terms to the front of the summations as constants. That is going to leave us with 1250 divided by N to the 4th power multiplied by the summation from 1 ton of I cubed. And for the second term, this is going to be -125 divided by N2d, multiplied by the summation from 1 ton of I. Now, for the summation of I cubed, the summation from 1 ton of I cubed is going to evaluate to the quantity and multiply by N + 1 divided by 2, that entire quantity squared. And in the second summation, the summation from I is equal 1 to 1 of N of just I is going to simplify to be N multiplied by M + 1 divided by 2. So by applying these substitutions, we will have 1250 divided by N to the fourth power, multiplied by the quantity, N times N + 1 divided by 2 quantities squared. Minus 125 N2 multiplied by the quantity, and multiplied by M + 1 divided by 2. By multiplying these terms together, we will be left with 1250 multiplied by M + 1 quantity squared, divided by 4 N squared. Minus 125, multiplied by N + 1 divided by 2n, and recall that by the definition, we still have a limit as N approaches infinity. By applying the infinite limit, this difference is going to simplify to be 1250 divided by 4 minus 125 divided by 2. This is going to simplify to be 312.5 minus 62.5, which is equal to 250, and this is going to be the value of the summation that we calculated earlier. And with that being said, the overall solution to this problem is going to be C. So I hope this video helps you in understanding how to approach this type of problem, and we will go ahead and see you all in the next video.

Limit definition of the definite integral Use the limit definition of the definite integral with right Riemann sums and a regular partition to evaluate the following definite integrals. Use the Fundamental Theorem of Calculus to check your answer.

∫₀⁴ (𝓍³―𝓍) d𝓍

Key Concepts

Definite Integral

Riemann Sums

Fundamental Theorem of Calculus

Watch next

Limit definition of the definite integral Use the limit definition of the definite integral with right Riemann sums and a regular partition to evaluate the following definite integrals. Use the Fundamental Theorem of Calculus to check your answer. ∫₀⁴ (𝓍³―𝓍) d𝓍

Key Concepts

Definite Integral

Riemann Sums

Fundamental Theorem of Calculus

Watch next

Limit definition of the definite integral Use the limit definition of the definite integral with right Riemann sums and a regular partition to evaluate the following definite integrals. Use the Fundamental Theorem of Calculus to check your answer.

∫₀⁴ (𝓍³―𝓍) d𝓍