Riemann problem - Introduction:
In math, a Riemann sum is a technique for similar to the sum area below a curve on a graph, or else identified as an integral. It may also be used to describe the integration operation. A function is definite to be Riemann integrable if the minor and higher Riemann sums get ever nearer as the partition obtain better and finer. This information can also be used for numerical integration.
Riemann Problem - Definition:
For a function f: S -> R, where S is a subset of the real numbers R,
I = [a, b] is a closed interval contained in S. A finite set of points `{x_0, x_1, x_2, ... x_n}` such that `a = x_0 < x_1 < x_2 ... < x_n = b` creates a partition `S = (x_0, x_1), (x_1, x_2), ... (x_(n-1), x_n] of I.`
Because T is a partition with n elements of I, the Riemann sum of f over I with the partition T is defined as
`M = \sum_{i=1}^{n} f(y_i)(x_{i}-x_{i-1})`
Where `x_(i-1) <= y_i <= x_i.`
The choice of `y_i` in this interval is arbitrary. If `y_i = x_(i-1)` for all i, then M is called a left Riemann sum. If `y_i = x_i` , then M is called a right Riemann sum. If `y_i = (x_i+x_(i-1))/2,` then M is called a middle Riemann sum. The average of the left Riemann sum and right Riemann sum is the trapezoidal sum.
Riemann Problem – Examples:
Riemann problem – Example 1:
`\int_{1}^{e^{\pi}} \frac{\sin (\ln x)}{x}\,dx\,`
Solution:
`u=\ln x\,`
`du=\frac{1}{x}dx\,`
So we have
`\int_{1}^{e^{\pi}} \frac{\sin (\ln x)}{x}\,dx=\int_0^{\pi}\sin u\,du\,`
Notice, the limits of integration changed because when `x=1\, and x=e^{\pi}\,, "we have "u=0\, and u=\pi`, respectively.
` \int_0^{\pi}\sin u\,du=-\cos u |_0^{\pi}=-\cos \pi+\cos 0=2\,`
Riemann problem – Example 2:
`\int \frac{x}{\sqrt{4+x^2}}\,dx\,`
Solution:
` u=4+x^2\,`
`du=2x\,dx\,`
` \int \frac{x}{\sqrt{4+x^2}}\,dx=\frac{1}{2}\int \frac{du}{\sqrt{u}}=u^{\frac{1}{2}}+C=\sqrt{x^2+4}+C\,`
I have recently faced lot of problem while learning Differentiation Rules, But thank to online resources of math which helped me to learn myself easily on net.
Riemann problem – Example 3:
`\int \frac{x}{1-x^2}\,dx\,`
Solution:
We could do this integral with partial fractions, but for instructive purposes let's use the method of trigonometric substitution.
`x=\sin \theta\,`
` dx=\cos \theta\,d\theta\,`
` \int \frac{x}{1-x^2}\,dx`
`=\int \frac{\sin \theta\cos \theta}{1-\sin^2\theta}d\theta`
`=\int \frac{\sin \theta\cos \theta}{\cos^2\theta}d\theta`
`=\int \frac{\sin \theta}{\cos \theta}d\theta\,`
At this point, we have to do a substitution again.
` u=\cos \theta\,`
`du=-\sin \theta\,d\theta\,`
` \int \frac{\sin \theta}{\cos \theta}d\theta=-\int \frac{du}{u}=-\ln |u|+C=-\ln |\cos \theta|+C\,`
Now, we know x = sinθ and we need to know the value of cosθ. Using the well known trig identity, sin2x + cos2x = 1, we get that` \cos \theta=\sqrt{1-x^2}` . So
` \int \frac{x}{1-x^2}\,dx=-\frac{1}{2}\ln |1-x^2|+C\,`
In math, a Riemann sum is a technique for similar to the sum area below a curve on a graph, or else identified as an integral. It may also be used to describe the integration operation. A function is definite to be Riemann integrable if the minor and higher Riemann sums get ever nearer as the partition obtain better and finer. This information can also be used for numerical integration.
Riemann Problem - Definition:
For a function f: S -> R, where S is a subset of the real numbers R,
I = [a, b] is a closed interval contained in S. A finite set of points `{x_0, x_1, x_2, ... x_n}` such that `a = x_0 < x_1 < x_2 ... < x_n = b` creates a partition `S = (x_0, x_1), (x_1, x_2), ... (x_(n-1), x_n] of I.`
Because T is a partition with n elements of I, the Riemann sum of f over I with the partition T is defined as
`M = \sum_{i=1}^{n} f(y_i)(x_{i}-x_{i-1})`
Where `x_(i-1) <= y_i <= x_i.`
The choice of `y_i` in this interval is arbitrary. If `y_i = x_(i-1)` for all i, then M is called a left Riemann sum. If `y_i = x_i` , then M is called a right Riemann sum. If `y_i = (x_i+x_(i-1))/2,` then M is called a middle Riemann sum. The average of the left Riemann sum and right Riemann sum is the trapezoidal sum.
Riemann Problem – Examples:
Riemann problem – Example 1:
`\int_{1}^{e^{\pi}} \frac{\sin (\ln x)}{x}\,dx\,`
Solution:
`u=\ln x\,`
`du=\frac{1}{x}dx\,`
So we have
`\int_{1}^{e^{\pi}} \frac{\sin (\ln x)}{x}\,dx=\int_0^{\pi}\sin u\,du\,`
Notice, the limits of integration changed because when `x=1\, and x=e^{\pi}\,, "we have "u=0\, and u=\pi`, respectively.
` \int_0^{\pi}\sin u\,du=-\cos u |_0^{\pi}=-\cos \pi+\cos 0=2\,`
Riemann problem – Example 2:
`\int \frac{x}{\sqrt{4+x^2}}\,dx\,`
Solution:
` u=4+x^2\,`
`du=2x\,dx\,`
` \int \frac{x}{\sqrt{4+x^2}}\,dx=\frac{1}{2}\int \frac{du}{\sqrt{u}}=u^{\frac{1}{2}}+C=\sqrt{x^2+4}+C\,`
I have recently faced lot of problem while learning Differentiation Rules, But thank to online resources of math which helped me to learn myself easily on net.
Riemann problem – Example 3:
`\int \frac{x}{1-x^2}\,dx\,`
Solution:
We could do this integral with partial fractions, but for instructive purposes let's use the method of trigonometric substitution.
`x=\sin \theta\,`
` dx=\cos \theta\,d\theta\,`
` \int \frac{x}{1-x^2}\,dx`
`=\int \frac{\sin \theta\cos \theta}{1-\sin^2\theta}d\theta`
`=\int \frac{\sin \theta\cos \theta}{\cos^2\theta}d\theta`
`=\int \frac{\sin \theta}{\cos \theta}d\theta\,`
At this point, we have to do a substitution again.
` u=\cos \theta\,`
`du=-\sin \theta\,d\theta\,`
` \int \frac{\sin \theta}{\cos \theta}d\theta=-\int \frac{du}{u}=-\ln |u|+C=-\ln |\cos \theta|+C\,`
Now, we know x = sinθ and we need to know the value of cosθ. Using the well known trig identity, sin2x + cos2x = 1, we get that` \cos \theta=\sqrt{1-x^2}` . So
` \int \frac{x}{1-x^2}\,dx=-\frac{1}{2}\ln |1-x^2|+C\,`
