Solve the following definite integrals using the substitution method: I) integral06 (6x)/(3x−2) dx II)…

Question

Solve the following definite integrals using the substitution method: I) integral06 (6x)/(3x−2) dx II) integral56 (1−2x)/(x−4) dx

ScanSolve AI Tutor · Accepted Answer

Here are the solutions for the definite integrals using the substitution method.

I) For the integral $\int_{0}^{6} \frac{6x}{3x-2}\,dx$ :

Define the substitution and find the differential.
Let $u = 3x-2$ .
Then $du = 3\,dx$ , which means $dx = \frac{1}{3}\,du$ .
Also, from $u = 3x-2$ , we have $3x = u+2$ , so $6x = 2(u+2)$ .
Change the limits of integration.
When $x=0$ , $u = 3(0)-2 = -2$ .
When $x=6$ , $u = 3(6)-2 = 18-2 = 16$ .
Substitute into the integral and simplify.
$\int_{-2}^{16} \frac{2(u+2)}{u} \cdot \frac{1}{3}\,du = \frac{2}{3} \int_{-2}^{16} \frac{u+2}{u}\,du = \frac{2}{3} \int_{-2}^{16} \left(1 + \frac{2}{u}\right)\,du$
Integrate and evaluate using the new limits.
$\frac{2}{3} \left[u + 2\ln|u|\right]_{-2}^{16}$ $\frac{2}{3} \left[(16 + 2\ln|16|) - (-2 + 2\ln|-2|)\right]$ $\frac{2}{3} \left[16 + 2\ln(16) + 2 - 2\ln(2)\right]$ $\frac{2}{3} \left[18 + 2(\ln(16) - \ln(2))\right]$ $\frac{2}{3} \left[18 + 2\ln\left(\frac{16}{2}\right)\right]$ $\frac{2}{3} \left[18 + 2\ln(8)\right]$ $\frac{2}{3} \cdot 2 \left[9 + \ln(8)\right]$ $\frac{4}{3} \left[9 + \ln(8)\right] = 12 + \frac{4}{3}\ln(8)$
Since $\ln(8) = \ln(2^3) = 3\ln(2)$ , the answer can also be written as:
$12 + \frac{4}{3}(3\ln(2)) = 12 + 4\ln(2)$
The final answer is $\boxed{12 + 4\ln(2)}$ .

II) For the integral $\int_{5}^{6} \frac{1-2x}{x-4}\,dx$ :
Define the substitution and find the differential.
Let $u = x-4$ .
Then $du = dx$ .
Also, from $u = x-4$ , we have $x = u+4$ . So $1-2x = 1-2(u+4) = 1-2u-8 = -2u-7$ .
Change the limits of integration.
When $x=5$ , $u = 5-4 = 1$ .
When $x=6$ , $u = 6-4 = 2$ .
Substitute into the integral and simplify.
$\int_{1}^{2} \frac{-2u-7}{u}\,du = \int_{1}^{2} \left(-2 - \frac{7}{u}\right)\,du$
Integrate and evaluate using the new limits.
$\left[-2u - 7\ln|u|\right]_{1}^{2}$ $\left[(-2(2) - 7\ln|2|) - (-2(1) - 7\ln|1|)\right]$ $\left[-4 - 7\ln(2)\right] - \left[-2 - 7(0)\right]$ $-4 - 7\ln(2) + 2$ $-2 - 7\ln(2)$
The final answer is $\boxed{-2 - 7\ln(2)}$ .

III) For the integral $\int_{0}^{3} \frac{2x-3}{2x+8}\,dx$ :
Define the substitution and find the differential.
Let $u = 2x+8$ .
Then $du = 2\,dx$ , which means $dx = \frac{1}{2}\,du$ .
Also, from $u = 2x+8$ , we have $2x = u-8$ . So $2x-3 = (u-8)-3 = u-11$ .
Change the limits of integration.
When $x=0$ , $u = 2(0)+8 = 8$ .
When $x=3$ , $u = 2(3)+8 = 6+8 = 14$ .
Substitute into the integral and simplify.
$\int_{8}^{14} \frac{u-11}{u} \cdot \frac{1}{2}\,du = \frac{1}{2} \int_{8}^{14} \left(1 - \frac{11}{u}\right)\,du$
Integrate and evaluate using the new limits.
$\frac{1}{2} \left[u - 11\ln|u|\right]_{8}^{14}$ $\frac{1}{2} \left[(14 - 11\ln|14|) - (8 - 11\ln|8|)\right]$ $\frac{1}{2} \left[14 - 11\ln(14) - 8 + 11\ln(8)\right]$ $\frac{1}{2} \left[6 + 11(\ln(8) - \ln(14))\right]$ $\frac{1}{2} \left[6 + 11\ln\left(\frac{8}{14}\right)\right]$ $\frac{1}{2} \left[6 + 11\ln\left(\frac{4}{7}\right)\right]$ $3 + \frac{11}{2}\ln\left(\frac{4}{7}\right)$
The final answer is $\boxed{3 + \frac{11}{2}\ln\left(\frac{4}{7}\right)}$ .

Solve the following definite integrals using the substitution method: I) integral06 (6x)/(3x-2)\,dx II) integral56 (1-2x)/(x-4)\,dx

Step-by-step solution

Define the substitution and find the differential.

Change the limits of integration.

Substitute into the integral and simplify.

Integrate and evaluate using the new limits.

Define the substitution and find the differential.

Change the limits of integration.

Substitute into the integral and simplify.

Integrate and evaluate using the new limits.

Define the substitution and find the differential.

Change the limits of integration.

Substitute into the integral and simplify.

Integrate and evaluate using the new limits.

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