Question

If \(\int x^3 e^x \, dx = e^x(px^3 + qx^2 + rx + s) + C\), then find the value of \(p + q + r + s\).

• A. \(-2\)
• B. \(0\)
• C. \(1\)
• D. \(6\)

Hint:

For integrals of the form \( \int x^n e^x dx \), use repeated differentiation of the polynomial (tabular or shortcut method) instead of full integration by parts each time.

Answer 1

The Correct Option is A

Concept: This type of integral, involving a polynomial multiplied by an exponential function, is solved using Integration by Parts repeatedly. The standard formula is: \[ \int u \, dv = uv - \int v \, du \] We choose \(u\) as the algebraic function because repeated differentiation eventually reduces it to zero (LIATE rule).

Step 1: First integration by parts.
Let \[ I = \int x^3 e^x \, dx \] Take: \[ u = x^3 \Rightarrow du = 3x^2 dx \] \[ dv = e^x dx \Rightarrow v = e^x \] Then: \[ I = x^3 e^x - 3\int x^2 e^x dx \quad \cdots (1) \]

Step 2: Second integration by parts.
For \( \int x^2 e^x dx \): \[ u = x^2,\quad du = 2x dx,\quad v = e^x \] \[ \int x^2 e^x dx = x^2 e^x - 2\int x e^x dx \] Substitute into (1): \[ I = x^3 e^x - 3(x^2 e^x - 2\int x e^x dx) \] \[ I = x^3 e^x - 3x^2 e^x + 6\int x e^x dx \quad \cdots (2) \]

Step 3: Third integration by parts.
For \( \int x e^x dx \): \[ u = x,\quad du = dx,\quad v = e^x \] \[ \int x e^x dx = x e^x - \int e^x dx = x e^x - e^x \] Substitute into (2): \[ I = x^3 e^x - 3x^2 e^x + 6(x e^x - e^x) + C \] \[ I = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C \]

Step 4: Factorization and comparison.
\[ I = e^x(x^3 - 3x^2 + 6x - 6) + C \] Comparing with: \[ I = e^x(px^3 + qx^2 + rx + s) + C \] We get: \[ p = 1,\quad q = -3,\quad r = 6,\quad s = -6 \]

Step 5: Required sum.
\[ p + q + r + s = 1 - 3 + 6 - 6 = -2 \]