23. Use the substitution u = √(x) to simplify the integral...

23. Use the substitution $u = \sqrt{x}$ to simplify the integral $\int_{x=1}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} dx$. (A) $2 \int_{u=1}^{16} e^u du$ (B) $2 \int_{u=1}^{4} e^u du$ (C) $\frac{1}{2} \int_{u=1}^{2} e^u du$ (D) $2 \int_{u=1}^{2} e^u du$ (E) None of these

Transcript

00:01 Let us consider an integral given by integral of 0 to 1 6 r divided by square root of 16 plus 3 r squared into dr.

00:14 We will make use of substitution to solve this integral.

00:17 So let us consider u to be equal to 16 plus 3r squared.

00:23 So we have d u to be equal to 6r into dr.

00:27 On substituting these values back in the integral, we will get integral.

00:31 Of 6 r d r will become d u divided by root u so that can be written as integral of u power minus 1 by 2 into d u.

00:46 So we will make use of the integration rule which is integral of x power n d x is equal to x power n plus 1 by n plus 1 plus c.

01:00 By making use of this formula here we will now obtain integration to be equal to u power minus 1 by 2 plus 1 by minus 1 by 2 plus 1 will give us u power 1 by 2 so we will obtain 2 by 1 into u power 1 by 2 so we have 2 u power 1 by 2 which is root u plus c therefore on substituting the value of u we have 2 into square root of 16 plus 3r square plus c.

01:47 So this is the value of the integral 6r d r divided by square root of 16 plus 3r square.

01:57 Now we will substitute the limits from 0 to 1.

02:01 So we have got to substitute the limits from 0 to 1.

02:06 So on substituting the limits we will have twice of root of 16 plus 3 is 19 and the lower limit will be minus root 16 my 0 will give us root 16 so we have twice of root 19 minus 4 so therefore we have now obtained the value of integral from 0 to 1 6 r d r divided by root of 16 plus 3 r square is to twice of root 19 minus 4.

02:49 Now the next part of the question is integral 0 to pi by 8 1 plus e power than 2x into sequence square 2x into d x.

03:09 Here let u be equal to 2x then d u will be equal to 2 dx.

03:16 On substituting we will first get integral 1 plus e power tan u into sikin square u and d x can be written as d u by u so we can put 1 by 2 outside and write du here.

03:37 Now let us again apply substitution so here let us consider let v to be equal to tan u then dv will be equal to second square u into d on substituting this data here we will get 1 by 2 1 plus e power v into dv now we will split the integral to these two terms and then integrate them so we will have 1 by 2 integral of dv plus 1 by 2 integral of e power v t v so we have 1 by 2 v plus e power v t v so we have 1 by 2 v plus e power v by 2.

04:16 So on substituting the value of v, we have 1 by 2 tan u and u is 2x plus e power tan 2x by 2 plus c.

04:32 So therefore we have now obtained integral of 1 plus e power tan 2x into secan square 2x into dx is equal to 1 by 2 tan 2x plus e .2 tan 2x plus e.

04:48 E power tan 2x divided by 2 plus c.

04:56 Now let us apply the limits which is from 0 to pi by 8.

05:00 So we have called limits from 0 to pi by 8...

Question

23. Use the substitution $u = \sqrt{x}$ to simplify the integral $\int_{x=1}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} dx$. (A) $2 \int_{u=1}^{16} e^u du$ (B) $2 \int_{u=1}^{4} e^u du$ (C) $\frac{1}{2} \int_{u=1}^{2} e^u du$ (D) $2 \int_{u=1}^{2} e^u du$ (E) None of these

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