1.
Question 1 \(\int x^7\,dx=\)
Step 1: Use the power rule for integration: \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\), where \(n\neq -1\). Step 2: Here \(n=7\). Step 3: Add \(1\) to the exponent: \(7+1=8\). Step 4: Divide by the new exponent: \(\int x^7\,dx=\frac{x^8}{8}+C\). Answer: B. \(\frac{x^8}{8}+C\)
2.
Question 2 \(\int x^{\frac{1}{3}}\,dx=\)
Step 1: Use the power rule: \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\). Step 2: Here \(n=\frac{1}{3}\). Step 3: Add \(1\) to the exponent: \(\frac{1}{3}+1=\frac{4}{3}\). Step 4: Divide by \(\frac{4}{3}\), which is the same as multiplying by \(\frac{3}{4}\). Step 5: Therefore, \(\int x^{\frac{1}{3}}\,dx=\frac{3}{4}x^{\frac{4}{3}}+C\). Answer: A. \(\frac{3}{4}x^{\frac{4}{3}}+C\)
3.
Question 3 \(\int 3\cos x\,dx=\)
Step 1: Factor out the constant \(3\): \(\int 3\cos x\,dx=3\int \cos x\,dx\). Step 2: Use the basic antiderivative \(\int \cos x\,dx=\sin x\). Step 3: Multiply by the constant: \(3\sin x+C\). Answer: D. \(3\sin x+C\)
4.
Question 4 \(\int e^{2x}\sqrt{e^{2x}-7}\,dx=\frac{2\left(e^{2x}-7\right)^{\frac{3}{2}}}{3}+C\).
Step 1: Use substitution. Let \(u=e^{2x}-7\). Step 2: Differentiate: \(du=2e^{2x}\,dx\), so \(e^{2x}\,dx=\frac{1}{2}du\). Step 3: Rewrite the integral: \(\int e^{2x}\sqrt{e^{2x}-7}\,dx=\frac{1}{2}\int u^{\frac{1}{2}}\,du\). Step 4: Integrate: \(\frac{1}{2}\cdot \frac{2}{3}u^{\frac{3}{2}}+C=\frac{u^{\frac{3}{2}}}{3}+C\). Step 5: Substitute back: \(\frac{\left(e^{2x}-7\right)^{\frac{3}{2}}}{3}+C\). Step 6: The given expression has an extra factor of \(2\), so it is false. Answer: A. False
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: Rewrite the integral: \(\int \frac{\cos(3x)}{\sqrt{\sin(3x)}}\,dx=\frac{1}{3}\int u^{-\frac{1}{2}}\,du\). Step 4: Integrate: \(\frac{1}{3}\cdot 2u^{\frac{1}{2}}+C=\frac{2}{3}\sqrt{u}+C\). Step 5: Substitute back: \(\frac{2}{3}\sqrt{\sin(3x)}+C\). Step 6: The given expression is missing the factor \(\frac{1}{3}\), so it is false. Answer: B. False
6.
Question 6 \(\int x^2\cos x\,dx=\)
Step 1: Use integration by parts with \(u=x^2\) and \(dv=\cos x\,dx\). Step 2: Then \(du=2x\,dx\) and \(v=\sin x\). Step 3: Apply \(\int u\,dv=uv-\int v\,du\): \(\int x^2\cos x\,dx=x^2\sin x-\int 2x\sin x\,dx\). Step 4: For \(\int x\sin x\,dx\), use integration by parts again with \(u=x\), \(dv=\sin x\,dx\), so \(du=dx\), \(v=-\cos x\). Step 5: Then \(\int x\sin x\,dx=-x\cos x+\sin x\). Step 6: Substitute this back: \(x^2\sin x-2(-x\cos x+\sin x)+C\). Step 7: Simplify: \(x^2\sin x+2x\cos x-2\sin x+C\). Answer: D. \(x^2\sin x+2x\cos x-2\sin x+C\)
7.
Question 7 \(\int_{0}^{1}x\sqrt{25+x^2}\,dx=\int_{0}^{1}\frac{\sqrt{u}}{2}\,du\), after doing the substitution given by \(u=25+x^2\).
Step 1: Use the substitution \(u=25+x^2\). Step 2: Differentiate: \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}du\). Step 3: Change the limits of integration. When \(x=0\), \(u=25+0^2=25\). When \(x=1\), \(u=25+1^2=26\). Step 4: The integral becomes \(\int_{25}^{26}\frac{\sqrt{u}}{2}\,du\), not \(\int_{0}^{1}\frac{\sqrt{u}}{2}\,du\). Step 5: The statement is false because the limits should change after substitution. Answer: B. False
8.
Question 8 \(\int_{-3}^{3}x^5\,dx=0\).
Step 1: Find the antiderivative: \(\int x^5\,dx=\frac{x^6}{6}\). Step 2: Evaluate from \(-3\) to \(3\): \(\left[\frac{x^6}{6}\right]_{-3}^{3}=\frac{3^6}{6}-\frac{(-3)^6}{6}\). Step 3: Since \((-3)^6=3^6\), the two values are equal. Step 4: Therefore, \(\frac{3^6}{6}-\frac{3^6}{6}=0\). Step 5: Also, \(x^5\) is an odd function, so its integral over \([-a,a]\) is \(0\). Answer: A. True
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