Question

Q2 Find the exact value of k such that $$\int_{0}^{\frac{1}{2k}} \frac{1}{\sqrt{1-k^2x^2}} dx = \int_{0}^{1} \ln(2x+1) dx.$$

Name: q2 find the exact value of k such that int0frac12k frac1sqrt1 k2x2 dx int01 ln2x1 dx 38966
Uploaded: 2024-04-28T03:54:36-08:00
Duration: 2 min 9 s
Channel: Andrew Noble
Description: q2 find the exact value of k such that int0frac12k frac1sqrt1 k2x2 dx int01 ln2x1 dx 38966

          Q2
Find the exact value of k such that
$$\int_{0}^{\frac{1}{2k}} \frac{1}{\sqrt{1-k^2x^2}} dx = \int_{0}^{1} \ln(2x+1) dx.$$

$q2 find the exact value of k such that int0frac12k frac1sqrt1 k2x2 dx int01 ln2x1 dx 38966$

Added by Brittney C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Andrew Noble

04/28/2024

Step 1

Let $u = kx$, then $du = k dx$, so $dx = \frac{1}{k} du$. When $x=0$, $u=0$. When $x=\frac{1}{2k}$, $u = k(\frac{1}{2k}) = \frac{1}{2}$. $$\int_{0}^{\frac{1}{2k}} \frac{1}{\sqrt{1-k^2x^2}} dx = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-u^2}} \frac{1}{k} du = Show more…

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