bc unit 6 (ab topics) quiz 2025\n(a) 6\n(b) 8\n(c) 10\n(d) 14\n(e) 18\n7. $int(e^{x}+e)dx=$\n(a)…

bc unit 6 (ab topics) quiz 2025\n(a) 6\n(b) 8\n(c) 10\n(d) 14\n(e) 18\n7. $int(e^{x}+e)dx=$\n(a) $e^{x}+c$\n(b) $2e^{x}+c$\n(c) $e^{x}+e + c$\n(d) $e^{x + 1}+ex + c$\n(e) $e^{x}+ex + c$\n8. if $int_{0}^{k}\frac{x}{x^{2}+4}dx=\frac{1}{2}ln4$, where $k>0$, then $k=$\n(a) 0\n(b) $sqrt{2}$\n(c) 2\n(d) $sqrt{12}$\n(e) $\frac{1}{2}\tan(lnsqrt{2})$\n9. $int_{1}^{e}(\frac{x^{2}-1}{x})dx=$\n(a) $e-\frac{1}{e}$\n(b) $e^{2}-e$\n(c) $\frac{e^{2}}{2}-e+\frac{1}{2}$\n(d) $e^{2}-2$\n(e) $\frac{e^{2}}{2}-\frac{3}{2}$

bc unit 6 (ab topics) quiz 2025\n(a) 6\n(b) 8\n(c) 10\n(d) 14\n(e) 18\n7. $int(e^{x}+e)dx=$\n(a) $e^{x}+c$\n(b) $2e^{x}+c$\n(c) $e^{x}+e + c$\n(d) $e^{x + 1}+ex + c$\n(e) $e^{x}+ex + c$\n8. if $int_{0}^{k}\frac{x}{x^{2}+4}dx=\frac{1}{2}ln4$, where $k>0$, then $k=$\n(a) 0\n(b) $sqrt{2}$\n(c) 2\n(d) $sqrt{12}$\n(e) $\frac{1}{2}\tan(lnsqrt{2})$\n9. $int_{1}^{e}(\frac{x^{2}-1}{x})dx=$\n(a) $e-\frac{1}{e}$\n(b) $e^{2}-e$\n(c) $\frac{e^{2}}{2}-e+\frac{1}{2}$\n(d) $e^{2}-2$\n(e) $\frac{e^{2}}{2}-\frac{3}{2}$

Answer

7.

Explanation:

Step1: Use integral sum - rule

$\int(e^{x}+e)dx=\int e^{x}dx+\int e dx$

Step2: Integrate each term

We know that $\int e^{x}dx = e^{x}+C_1$ and $\int e dx=ex + C_2$. Since $C_1 + C_2$ is just a constant $C$, the result is $e^{x}+ex + C$.

Answer:

(E) $e^{x}+ex + C$

8.

Explanation:

Step1: Use substitution

Let $u = x^{2}+4$, then $du = 2x dx$ and $x dx=\frac{1}{2}du$. When $x = 0$, $u = 4$; when $x = k$, $u=k^{2}+4$. So $\int_{0}^{k}\frac{x}{x^{2}+4}dx=\frac{1}{2}\int_{4}^{k^{2}+4}\frac{du}{u}$.

Step2: Integrate $\frac{1}{u}$

$\frac{1}{2}\int_{4}^{k^{2}+4}\frac{du}{u}=\frac{1}{2}[\ln u]_{4}^{k^{2}+4}=\frac{1}{2}(\ln(k^{2}+4)-\ln 4)$.

Step3: Set equal to given value

Since $\frac{1}{2}(\ln(k^{2}+4)-\ln 4)=\frac{1}{2}\ln 4$, then $\ln(k^{2}+4)-\ln 4=\ln 4$, $\ln(k^{2}+4)=2\ln 4=\ln 16$, so $k^{2}+4 = 16$, $k^{2}=12$, and $k=\sqrt{12}$ (because $k > 0$).

Answer:

(D) $\sqrt{12}$

9.

Explanation:

Step1: Simplify the integrand

$\frac{x^{2}-1}{x}=\frac{x^{2}}{x}-\frac{1}{x}=x-\frac{1}{x}$.

Step2: Integrate term - by - term

$\int_{1}^{e}(x - \frac{1}{x})dx=\int_{1}^{e}x dx-\int_{1}^{e}\frac{1}{x}dx$. We know that $\int x dx=\frac{1}{2}x^{2}+C$ and $\int\frac{1}{x}dx=\ln x + C$. So $\int_{1}^{e}x dx-\int_{1}^{e}\frac{1}{x}dx=[\frac{1}{2}x^{2}]{1}^{e}-[\ln x]{1}^{e}$.

Step3: Evaluate the definite integrals

$(\frac{1}{2}e^{2}-\frac{1}{2})-(\ln e-\ln 1)=\frac{e^{2}}{2}-\frac{1}{2}-(1 - 0)=\frac{e^{2}}{2}- \frac{3}{2}$.

Answer:

(E) $\frac{e^{2}}{2}-\frac{3}{2}$