Gemini vs Perplexity for Math

To solve the integral $\int x^3 e^x , dx$, we will use the method of Integration by Parts (IBP).

The formula for integration by parts is: $$\int u \, dv = uv - \int v \, du$$

The Strategy: We use the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to choose our $u$. Since $x^3$ is an algebraic function and $e^x$ is an exponential function, we will set $u$ to be the algebraic part ($x^3$). The goal is to differentiate the $x$ term until it becomes a constant. Because the power is $x^3$, we will need to go through exactly three cycles of integration by parts.

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Step-by-Step Solution

Cycle 1: Reducing $x^3$ to $x^2$

For our first pass, we assign:

• $u = x^3 \implies du = 3x^2 , dx$
• $dv = e^x , dx \implies v = e^x$

Applying the formula ($\int u , dv = uv - \int v , du$): $$\int x^3 e^x \, dx = x^3 e^x - \int 3x^2 e^x \, dx$$

Explanation of Cycle 1: We have successfully taken the original integral and produced a new term ($x^3 e^x$) minus a new integral ($\int 3x^2 e^x , dx$). The power of $x$ has been reduced from 3 to 2, bringing us closer to a solvable integral.

Cycle 2: Reducing $x^2$ to $x$

Now, we must apply integration by parts to the new integral: $\int 3x^2 e^x , dx$

• $u = 3x^2 \implies du = 6x , dx$
• $dv = e^x , dx \implies v = e^x$

Applying the formula to this part: $$\int 3x^2 e^x \, dx = 3x^2 e^x - \int 6x e^x \, dx$$

Now, we substitute this back into our primary equation. Be careful to distribute the negative sign: $$\int x^3 e^x \, dx = x^3 e^x - \left( 3x^2 e^x - \int 6x e^x \, dx \right)$$ $$\int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + \int 6x e^x \, dx$$

Explanation of Cycle 2: We reduced the power of the algebraic term again, from 2 to 1. Because of the minus sign in the IBP formula, substituting it back into the main equation flips the sign of the remaining integral to positive.

Cycle 3: Eliminating $x$ entirely

Finally, we apply integration by parts to our latest integral: $\int 6x e^x , dx$

• $u = 6x \implies du = 6 , dx$
• $dv = e^x , dx \implies v = e^x$

Applying the formula: $$\int 6x e^x \, dx = 6x e^x - \int 6 e^x \, dx$$

Substitute this back into the primary equation: $$\int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + \left( 6x e^x - \int 6 e^x \, dx \right)$$

Explanation of Cycle 3: We have finally reduced the algebraic term to a constant ($6$). The remaining integral, $\int 6 e^x , dx$, is trivial to solve because the integral of $e^x$ is simply $e^x$.

Final Evaluation:

Evaluate the last integral: $$\int 6 e^x \, dx = 6e^x$$

Put it all together and add the constant of integration ($C$): $$\int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C$$

To make it aesthetically pleasing, we can factor out the $e^x$: Final Answer: $e^x(x^3 - 3x^2 + 6x - 6) + C$

(Note: The alternating signs $+, -, +, -$ are a hallmark of repeated integration by parts, which is why the "Tabular Method" is often used as a shortcut for integrals like this).

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Verification by Differentiation

To prove that our integral is correct, we will take the derivative of our answer. If correct, we should get the original integrand ($x^3 e^x$) back.

To differentiate $f(x) = e^x(x^3 - 3x^2 + 6x - 6)$, we use the Product Rule: $\frac{d}{dx}[uv] = u'v + uv'$

Let:

• $u = e^x \implies u' = e^x$
• $v = x^3 - 3x^2 + 6x - 6 \implies v' = 3x^2 - 6x + 6$

Applying the Product Rule: $$f'(x) = [e^x] \cdot (x^3 - 3x^2 + 6x - 6) + [e^x] \cdot (3x^2 - 6x + 6)$$

Factor out the $e^x$: $$f'(x) = e^x \left[ (x^3 - 3x^2 + 6x - 6) + (3x^2 - 6x + 6) \right]$$

Now, watch the highly satisfying cancellation of terms inside the brackets:

• The $-3x^2$ and $+3x^2$ cancel to $0$.
• The $+6x$ and $-6x$ cancel to $0$.
• The $-6$ and $+6$ cancel to $0$.

We are left with: $$f'(x) = e^x(x^3)$$

This perfectly matches our original integral! The solution is verified to be correct.

The indefinite integral is $\int x^3 e^x \, dx = e^x (x^3 - 3x^2 + 6x - 6) + C$.[1]

Step-by-Step Solution Using Integration by Parts

Integration by parts applies the formula $\int u \, dv = uv - \int v \, du$, derived from the product rule.[1][6] For $\int x^3 e^x \, dx$, the integrand is a product of a polynomial (algebraic) and exponential; per the LIATE rule (prioritizing Algebraic for $u$ over Exponential for $dv$), set $u =$ polynomial to simplify via repeated differentiation.[1]

Cycle 1:

• Let $u = x^3$, so $du = 3x^2 \, dx$.
• Let $dv = e^x \, dx$, so $v = e^x$.
• Then $\int x^3 e^x \, dx = x^3 e^x - \int e^x (3x^2) \, dx = x^3 e^x - 3 \int x^2 e^x \, dx$.[1]

Cycle 2:

• For $\int x^2 e^x \, dx$: $u = x^2$, $du = 2x \, dx$; $dv = e^x \, dx$, $v = e^x$.
• $\int x^2 e^x \, dx = x^2 e^x - \int e^x (2x) \, dx = x^2 e^x - 2 \int x e^x \, dx$.
• Substitute back: $\int x^3 e^x \, dx = x^3 e^x - 3 \left[ x^2 e^x - 2 \int x e^x \, dx \right] = e^x (x^3 - 3x^2) + 6 \int x e^x \, dx$.[1]

Cycle 3:

• For $\int x e^x \, dx$: $u = x$, $du = dx$; $dv = e^x \, dx$, $v = e^x$.
• $\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x$.[1]
• Substitute back: $\int x^3 e^x \, dx = e^x (x^3 - 3x^2) + 6 (x e^x - e^x) + C = e^x (x^3 - 3x^2 + 6x - 6) + C$.

Each cycle reduces the polynomial degree by 1 (from $x^3$ to $x^2$ to $x$ to constant), making the new integral easier until it reaches $\int e^x \, dx = e^x$.[1][3]

Verification by Differentiation

Differentiate $e^x (x^3 - 3x^2 + 6x - 6)$ using the product rule: $\frac{d}{dx} [e^x f(x)] = e^x f(x) + e^x f'(x)$, where $f(x) = x^3 - 3x^2 + 6x - 6$.

• $f'(x) = 3x^2 - 6x + 6 - 6 = 3x^2 - 6x = 3x(x - 2)$. (Constant term vanishes.)
• Full derivative: $e^x (x^3 - 3x^2 + 6x - 6) + e^x (3x^2 - 6x) = e^x [x^3 - 3x^2 + 6x - 6 + 3x^2 - 6x] = e^x x^3$.

This matches the integrand, confirming the antiderivative.[1]

Tabular Method Alternative (Efficient for Polynomials)

For repeated cycles, use the tabular (DI) method: differentiate the polynomial column until 0, integrate the exponential column, and sum with alternating signs (diagonal products).[1][3]

Differentiate ($u$)	Sign	Integrate ($dv$)
$x^3$	+	$e^x$
$3x^2$	-	$e^x$
$6x$	+	$e^x$
$6$	-	$e^x$
0

Result: $+ (x^3 e^x) - (3x^2 e^x) + (6x e^x) - (6 e^x) + C = e^x (x^3 - 3x^2 + 6x - 6) + C$.[1][3]