Suppose the graph of consists of two regions between and : from to , forms a triangle above the -axis with area ; from to , forms a rectangle below the -axis with area . What is the value of the definite integral ?
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- 0. Functions7h 55m
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- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
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8. Definite Integrals
Introduction to Definite Integrals
Multiple Choice
2. Evaluate the definite integral from to : .
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Verified step by step guidance1
Step 1: Recognize that the integral involves the product of x and ln(x). This suggests using integration by parts, which is given by the formula: ∫u dv = uv - ∫v du.
Step 2: Choose u = ln(x) and dv = x dx. Then, compute du = (1/x) dx and v = (x^2)/2 (since the integral of x dx is (x^2)/2).
Step 3: Substitute into the integration by parts formula: ∫x ln(x) dx = uv - ∫v du = ln(x) * (x^2)/2 - ∫((x^2)/2) * (1/x) dx.
Step 4: Simplify the remaining integral: ∫((x^2)/2) * (1/x) dx = ∫x/2 dx = (1/2) * ∫x dx. The integral of x dx is (x^2)/2, so this becomes (1/2) * (x^2)/2 = (x^2)/4.
Step 5: Combine the results and evaluate the definite integral from 1 to e. Substitute the limits into the expression obtained after integration by parts: [(ln(x) * (x^2)/2 - (x^2)/4)] evaluated from x = 1 to x = e.
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