1.
Question 1. \(\int 3\cos x\,dx=\)
Step 1: Factor out the constant \(3\): \(\int 3\cos x\,dx=3\int\cos x\,dx\). Step 2: The antiderivative of \(\cos x\) is \(\sin x\). Step 3: Therefore, \(3\int\cos x\,dx=3\sin x+C\). Answer: \(3\sin x+C\)
2.
Question 2. \(\int\frac{x^5+x^4+x^3}{x^3}\,dx=\frac{2x^3}{3}+\frac{2x^2}{3}+2x+C\).
Step 1: Simplify the integrand first: \(\frac{x^5+x^4+x^3}{x^3}=x^2+x+1\). Step 2: Integrate term by term: \(\int(x^2+x+1)\,dx=\frac{x^3}{3}+\frac{x^2}{2}+x+C\). Step 3: The given answer is \(\frac{2x^3}{3}+\frac{2x^2}{3}+2x+C\), which does not match the correct antiderivative. Answer: False
3.
Question 3. \(\int(\sec^2x+\tan x\sec x)\,dx=?\)
Step 1: Split the integral: \(\int(\sec^2x+\tan x\sec x)\,dx=\int\sec^2x\,dx+\int\tan x\sec x\,dx\). Step 2: The antiderivative of \(\sec^2x\) is \(\tan x\). Step 3: The antiderivative of \(\tan x\sec x\) is \(\sec x\). Step 4: Add the results: \(\tan x+\sec x+C\). Answer: \(\tan x+\sec x+C\)
4.
Question 4. In order to solve \(\int x^5\cos(3x^6)\,dx\), a good choice for a substitution would be \(u=3x^6\).
Step 1: Look at the inside function of the cosine: \(3x^6\). Step 2: Let \(u=3x^6\). Step 3: Differentiate: \(du=18x^5\,dx\). Step 4: The integral contains \(x^5\,dx\), which matches the derivative of \(3x^6\) up to a constant factor. Step 5: Therefore, \(u=3x^6\) is a good substitution. Answer: True
5.
Question 5. \(\int\frac{\cos(3x)}{\sqrt{\sin(3x)}}\,dx=2\sqrt{\sin(3x)}+C\).
Step 1: Use substitution. Let \(u=\sin(3x)\). Step 2: Differentiate: \(du=3\cos(3x)\,dx\), so \(\cos(3x)\,dx=\frac{1}{3}du\). Step 3: Substitute into the integral: \(\int\frac{\cos(3x)}{\sqrt{\sin(3x)}}\,dx=\frac{1}{3}\int u^{-1/2}\,du\). Step 4: Integrate: \(\frac{1}{3}\cdot 2u^{1/2}+C=\frac{2}{3}\sqrt{u}+C\). Step 5: Substitute back: \(\frac{2}{3}\sqrt{\sin(3x)}+C\). Step 6: The given answer is missing the factor \(\frac{1}{3}\), so it is false. Answer: False
6.
Question 6. \(\int xe^x\,dx=?\)
Step 1: Use integration by parts: \(\int u\,dv=uv-\int v\,du\). Step 2: Let \(u=x\), so \(du=dx\). Let \(dv=e^x\,dx\), so \(v=e^x\). Step 3: Substitute into the formula: \(\int xe^x\,dx=xe^x-\int e^x\,dx\). Step 4: Integrate: \(\int e^x\,dx=e^x\). Step 5: Therefore, \(\int xe^x\,dx=xe^x-e^x+C\). Answer: \(xe^x-e^x+C\)
7.
Question 7. \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}2\cos(3x)\,dx=-2\).
Step 1: Find the antiderivative: \(\int2\cos(3x)\,dx=\frac{2}{3}\sin(3x)\). Step 2: Evaluate from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\): \(\frac{2}{3}\left[\sin\left(\frac{3\pi}{2}\right)-\sin\left(-\frac{3\pi}{2}\right)\right]\). Step 3: \(\sin\left(\frac{3\pi}{2}\right)=-1\) and \(\sin\left(-\frac{3\pi}{2}\right)=1\). Step 4: The value is \(\frac{2}{3}(-1-1)=\frac{2}{3}(-2)=-\frac{4}{3}\). Step 5: Since \(-\frac{4}{3}\ne -2\), the statement is false. Answer: False
8.
Question 8. \(\int_0^1\frac{x}{x-5}\,dx\approx -0.12\).
Step 1: Use algebraic division: \(\frac{x}{x-5}=1+\frac{5}{x-5}\). Step 2: Rewrite the integral: \(\int_0^1\frac{x}{x-5}\,dx=\int_0^1 1\,dx+\int_0^1\frac{5}{x-5}\,dx\). Step 3: Integrate: \(\int 1\,dx=x\) and \(\int\frac{5}{x-5}\,dx=5\ln|x-5|\). Step 4: Evaluate: \([x+5\ln|x-5|]_0^1\). Step 5: Substitute the bounds: \(1+5\ln4-(0+5\ln5)=1+5\ln\left(\frac{4}{5}\right)\). Step 6: Calculate: \(1+5\ln(0.8)\approx -0.1157\approx -0.12\). Answer: True
1 out of 1
