1.
Question 1 \(\int \frac{x^5+x^4+x^3}{x^3}\,dx=\frac{2x^3}{3}+\frac{2x^2}{3}+2+C\).
Step 1: Simplify the integrand: \(\frac{x^5+x^4+x^3}{x^3}=x^2+x+1\). Step 2: Integrate each term separately: \(\int (x^2+x+1)\,dx=\int x^2\,dx+\int x\,dx+\int 1\,dx\). Step 3: Apply the power rule: \(\int x^2\,dx=\frac{x^3}{3}\), \(\int x\,dx=\frac{x^2}{2}\), and \(\int 1\,dx=x\). Step 4: The correct antiderivative is \(\frac{x^3}{3}+\frac{x^2}{2}+x+C\), which is not the expression given. Answer: A. False
2.
Question 2 \(\int(3x^2+2x-5)\,dx=x^3+x^2-5x+C\).
Step 1: Integrate each term separately: \(\int(3x^2+2x-5)\,dx=\int 3x^2\,dx+\int 2x\,dx-\int 5\,dx\). Step 2: Apply the power rule: \(\int 3x^2\,dx=x^3\), \(\int 2x\,dx=x^2\), and \(\int 5\,dx=5x\). Step 3: Combine the results and add the constant of integration: \(x^3+x^2-5x+C\). Step 4: This matches the statement, so the statement is true. Answer: A. True
3.
Question 3 \(\int(\sec^2 x+\tan x\sec x)\,dx=\)
Step 1: Split the integral: \(\int(\sec^2 x+\tan x\sec x)\,dx=\int \sec^2 x\,dx+\int \tan x\sec x\,dx\). Step 2: Use the basic antiderivative \(\int \sec^2 x\,dx=\tan x\). Step 3: Use the basic antiderivative \(\int \sec x\tan x\,dx=\sec x\). Step 4: Combine the results: \(\tan x+\sec x+C\). Answer: B. \(\tan x+\sec x+C\)
4.
Question 4 \(\int(\sin^4 x\cos x)\,dx=\)
Step 1: Use substitution. Let \(u=\sin x\). Step 2: Then \(du=\cos x\,dx\). Step 3: Rewrite the integral: \(\int \sin^4 x\cos x\,dx=\int u^4\,du\). Step 4: Integrate: \(\int u^4\,du=\frac{u^5}{5}+C\). Step 5: Substitute back \(u=\sin x\): \(\frac{\sin^5 x}{5}+C\). Answer: B. \(\frac{\sin^5 x}{5}+C\)
5.
Question 5 \(\int \frac{1}{(x-3)^2}\,dx=\)
Step 1: Rewrite the integrand using a negative exponent: \(\frac{1}{(x-3)^2}=(x-3)^{-2}\). Step 2: Use substitution. Let \(u=x-3\), so \(du=dx\). Step 3: Integrate: \(\int u^{-2}\,du=\frac{u^{-1}}{-1}+C=-u^{-1}+C\). Step 4: Substitute back \(u=x-3\): \(-\frac{1}{x-3}+C\). Answer: A. \(-\frac{1}{x-3}+C\)
6.
Question 6 \(\int x\sin x\,dx=\)
Step 1: Use integration by parts: \(\int u\,dv=uv-\int v\,du\). Step 2: Choose \(u=x\), so \(du=dx\). Choose \(dv=\sin x\,dx\), so \(v=-\cos x\). Step 3: Substitute into the formula: \(\int x\sin x\,dx=x(-\cos x)-\int(-\cos x)\,dx\). Step 4: Simplify: \(-x\cos x+\int \cos x\,dx\). Step 5: Integrate: \(\int \cos x\,dx=\sin x\). Step 6: The result is \(-x\cos x+\sin x+C\). Answer: A. \(-x\cos x+\sin x+C\)
7.
Question 7 \(\int_{0}^{1}\frac{x}{x-5}\,dx\approx -0.12\).
Step 1: Rewrite the integrand by 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: Find the antiderivative: \(\int \left(1+\frac{5}{x-5}\right)dx=x+5\ln|x-5|\). Step 4: Evaluate from \(0\) to \(1\): \([x+5\ln|x-5|]_{0}^{1}=1+5\ln 4-5\ln 5\). Step 5: Simplify: \(1+5\ln\left(\frac{4}{5}\right)\approx -0.1157\approx -0.12\). Step 6: The statement is true. Answer: A. True
8.
Question 8 The integral \(\int_{-5}^{4}x^6\,dx\), when evaluated, gives a positive number.
Step 1: Find the antiderivative: \(\int x^6\,dx=\frac{x^7}{7}\). Step 2: Evaluate from \(-5\) to \(4\): \(\left[\frac{x^7}{7}\right]_{-5}^{4}=\frac{4^7}{7}-\frac{(-5)^7}{7}\). Step 3: Since \((-5)^7\) is negative, subtracting it makes the value positive: \(\frac{4^7}{7}+\frac{5^7}{7}\). Step 4: Also, \(x^6\ge 0\) for all \(x\), so its integral over the interval is positive. Answer: A. True
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