1.
Question 1. \(\int(3x^2+2x-5)\,dx=x^3+x^2-5x+C\).
Step 1: Integrate each term separately: \(\int(3x^2+2x-5)\,dx=\int3x^2\,dx+\int2x\,dx-\int5\,dx\). Step 2: Use the power rule: \(\int3x^2\,dx=x^3\). Step 3: \(\int2x\,dx=x^2\). Step 4: \(\int5\,dx=5x\). Step 5: Combine the results: \(x^3+x^2-5x+C\). Step 6: This matches the statement. Answer: True
2.
Question 2. \(\int x^{\frac{1}{3}}\,dx=?\)
Step 1: Use the power rule for integrals: \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\), where \(n\ne-1\). 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 the new exponent: \(\int x^{\frac{1}{3}}\,dx=\frac{x^{\frac{4}{3}}}{\frac{4}{3}}+C\). Step 5: Simplify: \(\frac{1}{\frac{4}{3}}=\frac{3}{4}\). Step 6: Therefore, the integral is \(\frac{3}{4}x^{\frac{4}{3}}+C\). Answer: \(\frac{3}{4}x^{\frac{4}{3}}+C\)
3.
Question 3. \(\int\frac{4}{x+1}\,dx=4\ln|x+1|+C\).
Step 1: Factor out the constant \(4\): \(\int\frac{4}{x+1}\,dx=4\int\frac{1}{x+1}\,dx\). Step 2: Use the rule \(\int\frac{1}{u}\,du=\ln|u|+C\). Step 3: Let \(u=x+1\), so \(du=dx\). Step 4: The integral becomes \(4\ln|x+1|+C\). Step 5: This matches the statement. Answer: True
4.
Question 4. \(\int3x^2(x^3+4)^7\,dx=\)
Step 1: Use substitution. Let \(u=x^3+4\). Step 2: Differentiate: \(du=3x^2\,dx\). Step 3: Substitute into the integral: \(\int3x^2(x^3+4)^7\,dx=\int u^7\,du\). Step 4: Integrate using the power rule: \(\int u^7\,du=\frac{u^8}{8}+C\). Step 5: Substitute back \(u=x^3+4\). Step 6: The result is \(\frac{(x^3+4)^8}{8}+C\). Answer: \(\frac{(x^3+4)^8}{8}+C\)
5.
Question 5. \(\int \sin^4 x\cos x\,dx=\)
Step 1: Use substitution. Let \(u=\sin x\). Step 2: Then \(du=\cos x\,dx\). Step 3: Substitute into the integral: \(\int\sin^4x\cos x\,dx=\int u^4\,du\). Step 4: Integrate: \(\int u^4\,du=\frac{u^5}{5}+C\). Step 5: Substitute back \(u=\sin x\). Step 6: The result is \(\frac{\sin^5x}{5}+C\). Answer: \(\frac{\sin^5x}{5}+C\)
6.
Question 6. \(\int x^2\cos x\,dx=?\)
Step 1: Use integration by parts: \(\int u\,dv=uv-\int v\,du\). Step 2: Let \(u=x^2\), so \(du=2x\,dx\). Let \(dv=\cos x\,dx\), so \(v=\sin x\). Step 3: Then \(\int x^2\cos x\,dx=x^2\sin x-\int2x\sin x\,dx\). Step 4: Use integration by parts again for \(\int x\sin x\,dx\). Let \(u=x\), so \(du=dx\). Let \(dv=\sin x\,dx\), so \(v=-\cos x\). Step 5: Then \(\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x\). Step 6: Substitute 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: \(x^2\sin x+2x\cos x-2\sin x+C\)
7.
Question 7. The integral \(\int_0^3\frac{2|x|}{e^{3x}+\sin x}\,dx\), when evaluated, gives a negative number.
Step 1: On the interval \([0,3]\), \(|x|\ge0\), so \(2|x|\ge0\). Step 2: Also \(e^{3x}>0\) for all \(x\), and on \([0,3]\), the denominator \(e^{3x}+\sin x\) is positive. Step 3: Therefore, the integrand \(\frac{2|x|}{e^{3x}+\sin x}\ge0\) on \([0,3]\). Step 4: The integral of a nonnegative function over an interval cannot be negative. Step 5: Therefore, the statement is false. Answer: False
8.
Question 8. \(\int_{-3}^{3}x^5\,dx=0\).
Step 1: Notice that \(x^5\) is an odd function because \((-x)^5=-x^5\). Step 2: The interval \([-3,3]\) is symmetric about \(0\). Step 3: The integral of an odd function over a symmetric interval \([-a,a]\) is \(0\). Step 4: We can also compute directly: \(\int x^5\,dx=\frac{x^6}{6}\). Step 5: Evaluate: \(\left[\frac{x^6}{6}\right]_{-3}^{3}=\frac{3^6}{6}-\frac{(-3)^6}{6}=0\). Step 6: Therefore, the statement is true. Answer: True
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