▶️ Answer/Explanation
Solution
Substitute \( u = 1 + x \), \( du = dx \), \( x = 0 \to u = 1 \), \( x = 8 \to u = 9 \).
Integral: \(\int_{1}^{9} \frac{du}{\sqrt{u}} = \int_{1}^{9} u^{-1/2} \, du = [2u^{1/2}]_{1}^{9}\).
Evaluate: \( 2(9^{1/2}) – 2(1^{1/2}) = 2 \cdot 3 – 2 \cdot 1 = 6 – 2 = 4 \).
✅ Answer: D
▶️ Answer/Explanation
Solution
Substitute \( u = 1 – x^2 \), \( du = -2x \, dx \), so \( 2x \, dx = -du \).
Limits: \( x = 0 \to u = 1 \), \( x = 1/2 \to u = 1 – (1/2)^2 = 1 – 1/4 = 3/4 \).
Integral: \(\int_{1}^{3/4} \frac{-du}{\sqrt{u}} = -\int_{1}^{3/4} u^{-1/2} \, du = -[2u^{1/2}]_{1}^{3/4}\).
Evaluate: \(- (2 \cdot (3/4)^{1/2} – 2 \cdot 1^{1/2}) = – (2 \cdot \sqrt{3}/2 – 2) = – (\sqrt{3} – 2) = 2 – \sqrt{3}\).
✅ Answer: E
▶️ Answer/Explanation
Solution
Recognize \(\int \frac{1}{1 + x^2} \, dx = \arctan x + C\).
Constant factor: \(\int \frac{5}{1 + x^2} \, dx = 5 \int \frac{1}{1 + x^2} \, dx = 5 \arctan x + C\).
✅ Answer: D
▶️ Answer/Explanation
Solution
Use integration by parts: \(\int u \, dv = uv – \int v \, du\).
Let \( u = x \), \( dv = e^{-x} \, dx \), so \( du = dx \), \( v = \int e^{-x} \, dx = -e^{-x} \).
Integral: \(\int_{0}^{1} x e^{-x} \, dx = [x (-e^{-x})]_{0}^{1} – \int_{0}^{1} (-e^{-x}) \, dx\).
First term: \([-x e^{-x}]_{0}^{1} = (-1 \cdot e^{-1}) – (0 \cdot e^{0}) = -e^{-1}\).
Second term: \(\int_{0}^{1} e^{-x} \, dx = [-e^{-x}]_{0}^{1} = -e^{-1} – (-e^{0}) = -e^{-1} + 1\).
Total: \(-e^{-1} – (-e^{-1} + 1) = -e^{-1} + e^{-1} – 1 = 1 – 2e^{-1}\).
✅ Answer: C
▶️ Answer/Explanation
Solution
Substitute \( u = x^3 + 1 \), so \( du = 3x^2 \, dx \).
Integral: \(\int \frac{3x^2}{x^3 + 1} \, dx = \int \frac{du}{u} = \ln (x^3 + 1) + C\).
✅ Answer: D