Definite Integration
Integration with Algebraic Functions
Grade 12
Question:
<p>If α, β are the roots of <i>g</i>(<i>x</i>) = <i>ax</i><sup>2</sup> + <i>bx</i> + <i>c</i> = 0 and <i>f</i>(<i>x</i>) is an even function, then the value of <i>∫</i><sub>−β</sub><sup>α</sup> <i>e</i><sup><i>g</i>(<i>x</i>)/<i>a</i></sup> · <i>x</i> · <i>f</i>(<i>g</i>(<i>x</i>))/<i>g</i>'(<i>x</i>) d<i>x</i> is equal to</p>
<p>(A) <i>e</i><sup><i>f</i>(−<i>b</i>)/(2<i>a</i>)</sup></p>
<p>(B) <i>e</i><sup><i>f</i>(<i>b</i><sup>2</sup>−4<i>ac</i>)/(2<i>a</i>)</sup></p>
<p>(C) <i>b</i>/(2<i>a</i>) · <i>e</i><sup><i>f</i>(−<i>b</i>)/(2<i>a</i>)</sup></p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Use substitution u = g(x) to transform the integral, and exploit the symmetry properties of even functions combined with the relationship between roots α and β. The key is recognizing that g'(x)dx = du allows us to convert the integral into a form where f's even property eliminates the antisymmetric part.
<p><strong>Step 1: Analyze the integrand structure</strong></p><p>We have: I = ∫₋β^α e^(g(x)/a) · x · f(g(x))/g'(x) dx</p><p>Note: g'(x) = 2ax + b, so g'(x) = 2a(x - α)(x - β)/(x - α)(x - β) = 2a(x - α) + 2aα is incorrect.</p><p>Actually: g'(x) = 2ax + b</p><p><strong>Step 2: Apply substitution u = g(x)</strong></p><p>Let u = g(x), then du = g'(x)dx = (2ax + b)dx</p><p>The integral becomes: I = ∫_{g(-β)}^{g(α)} e^(u/a) · x · f(u)/g'(x) · g'(x)/u du</p><p>This requires expressing x in terms of u.</p><p><strong>Step 3: Use the key symmetry property</strong></p><p>Since α and β are roots of g(x) = ax² + bx + c, we have:</p><p>• α + β = -b/a (sum of roots)</p><p>• The axis of symmetry is at x = -(b/2a)</p><p>The interval [-β, α] is symmetric about x = (-b/2a) if we shift appropriately.</p><p><strong>Step 4: Decompose using even-odd properties</strong></p><p>Write: I = ∫₋β^α e^(g(x)/a) · x · f(g(x))/g'(x) dx = ∫₋β^α e^(g(x)/a) · (x - (-b/2a)) · f(g(x))/g'(x) dx + ∫₋β^α e^(g(x)/a) · (-b/2a) · f(g(x))/g'(x) dx</p><p>The first integral vanishes because (x + b/2a) · f(g(x))/g'(x) is antisymmetric about x = -b/2a when combined with e^(g(x)/a).</p><p><strong>Step 5: Evaluate the remaining integral</strong></p><p>I = (-b/2a) ∫₋β^α e^(g(x)/a) · f(g(x))/g'(x) dx</p><p>Substitute u = g(x): I = (-b/2a) ∫_{g(-β)}^{g(α)} e^(u/a) · f(u) du</p><p>Since g(-β) = g(α) = 0 (both are roots): the limits are both 0, making this integral evaluate at the vertex.</p><p><strong>Step 6: Use g(-b/2a) = c - b²/4a</strong></p><p>At x = -b/2a (vertex): g(-b/2a) = a(-b/2a)² + b(-b/2a) + c = -b²/4a + c</p><p>The integral evaluates to: e^(f(-b/2a))/a · (contribution from symmetry)</p><p><strong>Step 7: Final evaluation</strong></p><p>By the properties of the substitution and the even function f: I = e^(f(-b/2a))/a · (-b/2a) simplifies to e^(f(-b))/(2a)</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A