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Find the value of 
int_(6)^(8)(2dx)/(4-x). Write your answer as the logarithm of a single number in simplest form.
Answer: 
ln(◻)

Find the value of 682dx4x \int_{6}^{8} \frac{2 d x}{4-x} . Write your answer as the logarithm of a single number in simplest form.\newlineAnswer: ln() \ln (\square)

Full solution

Q. Find the value of 682dx4x \int_{6}^{8} \frac{2 d x}{4-x} . Write your answer as the logarithm of a single number in simplest form.\newlineAnswer: ln() \ln (\square)
  1. Substitution Method: To solve this integral, we can use a substitution method. Let's let u=4xu = 4 - x, which means du=dxdu = -dx. We need to adjust the integral to account for this substitution.
  2. Adjusting Integral: Now we substitute uu into the integral and adjust the limits of integration. When x=6x = 6, u=46=2u = 4 - 6 = -2. When x=8x = 8, u=48=4u = 4 - 8 = -4. Also, we need to substitute du-du for dxdx, which will change the sign of the integral.\newlineThe integral becomes:\newline242duu\int_{-2}^{-4} \frac{-2du}{u}
  3. Integrating with Respect to u: We can now integrate with respect to uu. The integral of 2u-\frac{2}{u} dudu is 2lnu-2\ln|u|. We will evaluate this from 2-2 to 4-4.
  4. Evaluating Antiderivative: Evaluating the antiderivative at the bounds gives us:\newline2ln4+2ln2-2\ln|-4| + 2\ln|-2|
  5. Simplifying Expression: Simplify the expression using properties of logarithms. Since the logarithm of a negative number is not defined in the real number system, we can remove the negative signs inside the logarithms because lna=lna\ln|-a| = \ln|a| for any positive aa. This simplifies to: 2ln(4)+2ln(2)-2\ln(4) + 2\ln(2)
  6. Further Simplification: Now we use the property of logarithms that ln(ab)=bln(a)\ln(a^b) = b\cdot\ln(a) to further simplify the expression.\newlineThis gives us:\newline2ln(22)+2ln(2)-2\ln(2^2) + 2\ln(2)
  7. Combining Like Terms: Simplify the expression by applying the power rule for logarithms and combining like terms.\newlineThis simplifies to:\newline4ln(2)+2ln(2)-4\ln(2) + 2\ln(2)
  8. Final Answer: Combine the logarithmic terms to get the final answer.\newlineThis gives us:\newline2ln(2)-2\ln(2)
  9. Final Answer: Combine the logarithmic terms to get the final answer.\newlineThis gives us:\newline2ln(2)-2\ln(2)We can now express the final answer as the logarithm of a single number.\newlineThe final answer is:\newlineln(22)\ln(2^{-2})