Question:
If \( \int e^{5x}x^{n} \, dx = F(n, x) + c \), then \( 5F(n, x) + nF(n-1, x) = \)
If \( \int e^{5x}x^{n} \, dx = F(n, x) + c \), then \( 5F(n, x) + nF(n-1, x) = \)
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Reduction formulas are just structural rearrangements of integration by parts. Recognizing that the derivative of the integral equals the integrand lets you skip solving the reduction loop entirely.
Updated On: Jun 8, 2026
- \( F'(n, x) + k \)
- \( -F'(n, x) + k \)
- \( \frac{xF'(n, x)}{5} + k \)
- \( \frac{x^{2}F'(n, x)}{F(n, x)} + k \)
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The Correct Option is A
Solution and Explanation
Concept:
According to the fundamental theorem of calculus, differentiating an integral function with respect to \( x \) yields the integrand function itself:
\[
\frac{d}{dx}[F(n, x)] = F'(n, x) = e^{5x}x^n
\]
Step 1: Applying integration by parts to set up a reduction formula.
Let us perform integration by parts on the main integral expression \( I_n = \int e^{5x}x^n \, dx \). Let \( u = x^n \implies du = n x^{n-1} dx \) and \( dv = e^{5x} dx \implies v = \frac{e^{5x}}{5} \): \[ F(n, x) = x^n \cdot \frac{e^{5x}}{5} - \int \frac{e^{5x}}{5} \cdot n x^{n-1} dx \] \[ F(n, x) = \frac{1}{5}e^{5x}x^n - \frac{n}{5}F(n-1, x) \]
Step 2: Rearranging the terms to match the target expression.
Multiply the entire reduction equation by 5: \[ 5F(n, x) = e^{5x}x^n - nF(n-1, x) \] Bringing the negative term to the left side: \[ 5F(n, x) + nF(n-1, x) = e^{5x}x^n \]
Step 3: Expressing the right side in terms of derivative functions.
As established by our core concept definition, \( e^{5x}x^n = F'(n, x) \). Therefore: \[ 5F(n, x) + nF(n-1, x) = F'(n, x) \] This matches option (A) perfectly (where \( k \) represents any matching boundary shift constant).
Step 1: Applying integration by parts to set up a reduction formula.
Let us perform integration by parts on the main integral expression \( I_n = \int e^{5x}x^n \, dx \). Let \( u = x^n \implies du = n x^{n-1} dx \) and \( dv = e^{5x} dx \implies v = \frac{e^{5x}}{5} \): \[ F(n, x) = x^n \cdot \frac{e^{5x}}{5} - \int \frac{e^{5x}}{5} \cdot n x^{n-1} dx \] \[ F(n, x) = \frac{1}{5}e^{5x}x^n - \frac{n}{5}F(n-1, x) \]
Step 2: Rearranging the terms to match the target expression.
Multiply the entire reduction equation by 5: \[ 5F(n, x) = e^{5x}x^n - nF(n-1, x) \] Bringing the negative term to the left side: \[ 5F(n, x) + nF(n-1, x) = e^{5x}x^n \]
Step 3: Expressing the right side in terms of derivative functions.
As established by our core concept definition, \( e^{5x}x^n = F'(n, x) \). Therefore: \[ 5F(n, x) + nF(n-1, x) = F'(n, x) \] This matches option (A) perfectly (where \( k \) represents any matching boundary shift constant).
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