Which of the following integrals are improper?∫4515x−1&nbsp;dx&nbsp;&nbsp;х∫0112х−1&nbsp;dx&nbsp;&nbsp;∫∞−∞sin⁡(x)1+3x2&nbsp;dx&nbsp;&nbsp;∫13ln⁡(x−1)&nbsp;dx&nbsp;

Question

Which of the following integrals are improper?∫4515x−1&amp;nbsp;dx&amp;nbsp;&amp;nbsp;х∫0112х−1&amp;nbsp;dx&amp;nbsp;&amp;nbsp;∫∞−∞sin⁡(x)1+3x2&amp;nbsp;dx&amp;nbsp;&amp;nbsp;∫13ln⁡(x−1)&amp;nbsp;dx&amp;nbsp;

John Twitchell · Accepted Answer

A definite integral with one or both infinite bounds is referred to as an improper integral, as is an integrand that approaches infinity at one or more places in the integration range.1)∫4515x−1∫1x=ln|x|∫4515x−1=15ln|5x−1|∣45=15{ln|24|−ln|19|}Integral is not improper.2)∫0112x−1plugging&amp;nbsp;x=12,the&amp;nbsp;integral&amp;nbsp;appto&amp;nbsp;aches→0,&amp;nbsp;and&amp;nbsp;x=12&amp;nbsp;is within the range of integration [0,1].So this is an improper integration.3)∫−∞∞sin⁡x1+3x2&amp;nbsp;dx&amp;nbsp;It is an improper integral.4)∫13ln⁡(x−1)&amp;nbsp;dx&amp;nbsp;at x=1, the integral becomes undefined and x=1 is within the range of integration.so it is an improper integration.Answer that option (2),(3) and (4) are examples of improper integration.

star233 · Accepted Answer

Answer:a) The integral ∫4515x−1dx is not improper.b) The integral x∫0112x−1dx is improper.c) The integral ∫−∞∞sin(x)1+3x2dx is improper.d) The integral ∫13ln(x−1)dx is not improper.Explanation:a) To determine if the integral ∫4515x−1dx is improper, we need to check if the interval of integration includes any points of discontinuity or if the integrand becomes unbounded within that interval. In this case, the integrand 15x−1 is continuous and defined for all x within the interval [4,5]. Therefore, the integral is not improper.b) The integral x∫0112x−1dx is improper because the integrand 12x−1 becomes unbounded when x=12, which falls within the interval of integration [0,1]. To evaluate this improper integral, we split it into two parts:x∫0112x−1dx=lima→12−(x∫0a12x−1dx)+limb→12+(x∫b112x−1dx)c) The integral ∫−∞∞sin(x)1+3x2dx is improper because the interval of integration extends to infinity. To evaluate this improper integral, we can use the symmetry property of the sine function:∫−∞∞sin(x)1+3x2dx=2∫0∞sin(x)1+3x2dxd) The integral ∫13ln(x−1)dx is not improper because both the integrand ln(x−1) and the interval of integration [1,3] are well-defined and finite.

alenahelenash · Accepted Answer

Step 1:a) ∫4515x−1dx is not improper because the limits of integration, 4 and 5, are finite.Step 2:b) x∫0112x−1dx is improper because the integrand becomes infinite at x=12.Step 3:c) ∫−∞∞sin(x)1+3x2dx is improper because the limits of integration are infinite.Step 4:d) ∫13ln(x−1)dx is not improper because the limits of integration, 1 and 3, are finite.

karton · Accepted Answer

To determine which of the following integrals are improper, let&#039;s analyze each integral one by one.1) ∫4515x−1dxThis integral is a definite integral over a finite interval, from x=4 to x=5. Since the interval is finite, this integral is not improper.2) x∫0112x−1dxSimilar to the previous integral, this is also a definite integral over a finite interval, from x=0 to x=1. Thus, this integral is not improper.3) ∫−∞∞sin(x)1+3x2dxThis integral is an example of an improper integral because the integration limits extend to infinity. To solve this integral, we can split it into two separate integrals:∫−∞∞sin(x)1+3x2dx=∫−∞0sin(x)1+3x2dx+∫0∞sin(x)1+3x2dxEach of these individual integrals is improper, and we need to evaluate them separately.4) ∫13ln(x−1)dxThis integral is a definite integral over a finite interval, from x=1 to x=3. Therefore, it is not an improper integral.In summary:- The integrals in parts 1) and 2) are not improper integrals.- The integral in part 3) is an improper integral.- The integral in part 4) is not an improper integral.

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