Question

$\int e^x(x^3-2x^2+3x-4)dx =$

• A. $-e^x(x^3-x^2+5x-1)+c$
• B. $e^x(x^3-x^2+5x-1)+c$
• C. $e^x(x^3+x^2+5x+1)+c$
• D. $-e^x(x^3+x^2+5x+1)+c$
• E. None of these

Hint:

For integrals of the form $\int e^{ax}P(x)dx$, you can use the tabular integration method (a shortcut for repeated integration by parts) or use the formula $\int e^x P(x)dx = e^x(P(x)-P'(x)+P''(x)-...)+C$. In this case, $P(x)=x^3-2x^2+3x-4$, $P'(x)=3x^2-4x+3$, $P''(x)=6x-4$, $P'''(x)=6$. The integral is $e^x((x^3-2x^2+3x-4)-(3x^2-4x+3)+(6x-4)-6)+C = e^x(x^3-5x^2+13x-17)+C$.

Answer 1

Answer: (E)

Step 1: Formula for Integral Using repeated integration by parts for \( \int e^{-x} P(x) dx \): \[ \int e^{-x} P(x) dx = -e^{-x} [ P(x) + P'(x) + P''(x) + P'''(x) + \dots ] \] Here \( P(x) = x^3 - 2x^2 + 3x - 4 \).
Step 2: Calculate Derivatives \( P(x) = x^3 - 2x^2 + 3x - 4 \) \( P'(x) = 3x^2 - 4x + 3 \) \( P''(x) = 6x - 4 \) \( P'''(x) = 6 \) \( P^{(4)}(x) = 0 \)
Step 3: Sum the terms Sum \( S = P(x) + P'(x) + P''(x) + P'''(x) \) \( S = (x^3 - 2x^2 + 3x - 4) + (3x^2 - 4x + 3) + (6x - 4) + 6 \) Group by powers of x: \( x^3 \) term: \( x^3 \) \( x^2 \) term: \( -2x^2 + 3x^2 = x^2 \) \( x \) term: \( 3x - 4x + 6x = 5x \) Constant term: \( -4 + 3 - 4 + 6 = 1 \) So \( S = x^3 + x^2 + 5x + 1 \).
Step 4: Final Result Integral is \( -e^{-x}(x^3 + x^2 + 5x + 1) + c \).