Grok vs Kimi for Math

Step-by-Step Solution: ∫ x³ e^{2x} dx

To solve ∫ x³ e^{2x} dx, we use integration by parts repeatedly. The formula is:

∫ u dv = u v − ∫ v du

We choose u as the polynomial (which decreases in degree with each differentiation) and dv as the exponential (easy to integrate). Don't forget the constant of integration +C at the end.

Step 1: First Application of Integration by Parts

Let u = x³, so du = 3x² dx.
Let dv = e^{2x} dx, so v = (1/2) e^{2x}.

Then:
∫ x³ e^{2x} dx = (x³)(1/2 e^{2x}) − ∫ (1/2 e^{2x})(3x² dx)
= (x³ / 2) e^{2x} − (3/2) ∫ x² e^{2x} dx.

Step 2: Second Application (on ∫ x² e^{2x} dx)

Let u = x², so du = 2x dx.
Let dv = e^{2x} dx, so v = (1/2) e^{2x}.

Then:
∫ x² e^{2x} dx = (x²)(1/2 e^{2x}) − ∫ (1/2 e^{2x})(2x dx)
= (x² / 2) e^{2x} − ∫ x e^{2x} dx.

Step 3: Third Application (on ∫ x e^{2x} dx)

Let u = x, so du = dx.
Let dv = e^{2x} dx, so v = (1/2) e^{2x}.

Then:
∫ x e^{2x} dx = (x)(1/2 e^{2x}) − ∫ (1/2 e^{2x}) dx
= (x / 2) e^{2x} − (1/2) ∫ e^{2x} dx
= (x / 2) e^{2x} − (1/2)(1/2 e^{2x})
= e^{2x} (x/2 − 1/4).

Step 4: Back-Substitute into Step 2

∫ x² e^{2x} dx = (x² / 2) e^{2x} − e^{2x} (x/2 − 1/4)
= e^{2x} [x²/2 − (x/2 − 1/4)]
= e^{2x} (x²/2 − x/2 + 1/4).

Step 5: Back-Substitute into Step 1 (Original Integral)

∫ x³ e^{2x} dx = (x³ / 2) e^{2x} − (3/2) [e^{2x} (x²/2 − x/2 + 1/4)]
= e^{2x} [x³/2 − (3/2)(x²/2 − x/2 + 1/4)]
= e^{2x} [x³/2 − (3/4)x² + (3/4)x − 3/8] + C.

Final Answer:
∫ x³ e^{2x} dx = e^{2x} \left( \frac{1}{2} x^3 - \frac{3}{4} x^2 + \frac{3}{4} x - \frac{3}{8} \right) + C.

Verification by Differentiation

Let y = e^{2x} \left( \frac{1}{2} x^3 - \frac{3}{4} x^2 + \frac{3}{4} x - \frac{3}{8} \right).
Differentiate using the product rule: y' = (e^{2x})' ⋅ (polynomial) + e^{2x} ⋅ (polynomial)'.

• Let p(x) = \frac{1}{2} x^3 - \frac{3}{4} x^2 + \frac{3}{4} x - \frac{3}{8}.
• p'(x) = \frac{3}{2} x^2 - \frac{3}{2} x + \frac{3}{4}.
• (e^{2x})' = 2 e^{2x}, so 2 p(x) = x^3 - \frac{3}{2} x^2 + \frac{3}{2} x - \frac{3}{4}.

Now:
y' = e^{2x} \left[ x^3 - \frac{3}{2} x^2 + \frac{3}{2} x - \frac{3}{4} \right] + e^{2x} \left[ \frac{3}{2} x^2 - \frac{3}{2} x + \frac{3}{4} \right]
= e^{2x} \left[ x^3 + \left( -\frac{3}{2} x^2 + \frac{3}{2} x^2 \right) + \left( \frac{3}{2} x - \frac{3}{2} x \right) + \left( -\frac{3}{4} + \frac{3}{4} \right) \right]
= e^{2x} (x^3) = x^3 e^{2x}.

The derivative matches the integrand, confirming the solution is correct.

To solve $\displaystyle \int x^3 e^{2x},dx$, we apply integration by parts three times, using the formula: $$\int u\,dv = uv - \int v\,du$$

The strategy is to differentiate the polynomial part ($x^3$) until it becomes a constant, while integrating the exponential part ($e^{2x}$) each time.

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First Application

Let:

• $u = x^3 \quad \Rightarrow \quad du = 3x^2,dx$
• $dv = e^{2x},dx \quad \Rightarrow \quad v = \frac{1}{2}e^{2x}$

Applying the formula: $$\int x^3 e^{2x}\,dx = \frac{1}{2}x^3 e^{2x} - \int \frac{1}{2}e^{2x} \cdot 3x^2\,dx$$

Simplifying: $$\int x^3 e^{2x}\,dx = \frac{1}{2}x^3 e^{2x} - \frac{3}{2}\int x^2 e^{2x}\,dx \tag{1}$$

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Second Application

Now we evaluate $\displaystyle \int x^2 e^{2x},dx$.

Let:

• $u = x^2 \quad \Rightarrow \quad du = 2x,dx$
• $dv = e^{2x},dx \quad \Rightarrow \quad v = \frac{1}{2}e^{2x}$

Applying the formula: $$\int x^2 e^{2x}\,dx = \frac{1}{2}x^2 e^{2x} - \int \frac{1}{2}e^{2x} \cdot 2x\,dx = \frac{1}{2}x^2 e^{2x} - \int x e^{2x}\,dx$$

Substituting back into equation (1): $$\int x^3 e^{2x}\,dx = \frac{1}{2}x^3 e^{2x} - \frac{3}{2}\left(\frac{1}{2}x^2 e^{2x} - \int x e^{2x}\,dx\right)$$

Distributing the $-\frac{3}{2}$: $$\int x^3 e^{2x}\,dx = \frac{1}{2}x^3 e^{2x} - \frac{3}{4}x^2 e^{2x} + \frac{3}{2}\int x e^{2x}\,dx \tag{2}$$

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Third Application

Now we evaluate $\displaystyle \int x e^{2x},dx$.

Let:

• $u = x \quad \Rightarrow \quad du = dx$
• $dv = e^{2x},dx \quad \Rightarrow \quad v = \frac{1}{2}e^{2x}$

Applying the formula: $$\int x e^{2x}\,dx = \frac{1}{2}x e^{2x} - \int \frac{1}{2}e^{2x}\,dx = \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x}$$

Substituting back into equation (2): $$\int x^3 e^{2x}\,dx = \frac{1}{2}x^3 e^{2x} - \frac{3}{4}x^2 e^{2x} + \frac{3}{2}\left(\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x}\right) + C$$

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Final Simplification

Distribute and combine terms: $$\int x^3 e^{2x}\,dx = \frac{1}{2}x^3 e^{2x} - \frac{3}{4}x^2 e^{2x} + \frac{3}{4}x e^{2x} - \frac{3}{8}e^{2x} + C$$

Factor out $e^{2x}$: $$= e^{2x}\left(\frac{1}{2}x^3 - \frac{3}{4}x^2 + \frac{3}{4}x - \frac{3}{8}\right) + C$$

With a common denominator of 8: $$\boxed{\int x^3 e^{2x}\,dx = \frac{e^{2x}}{8}\left(4x^3 - 6x^2 + 6x - 3\right) + C}$$

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Verification by Differentiation

Let $F(x) = \dfrac{e^{2x}}{8}(4x^3 - 6x^2 + 6x - 3)$. Using the product rule: $$F'(x) = \frac{2e^{2x}}{8}(4x^3 - 6x^2 + 6x - 3) + \frac{e^{2x}}{8}(12x^2 - 12x + 6)$$

Factor out $\dfrac{e^{2x}}{8}$: $$F'(x) = \frac{e^{2x}}{8}\left[2(4x^3 - 6x^2 + 6x - 3) + (12x^2 - 12x + 6)\right]$$

Expand the brackets: $$F'(x) = \frac{e^{2x}}{8}\left[8x^3 - 12x^2 + 12x - 6 + 12x^2 - 12x + 6\right]$$

Combine like terms: $$F'(x) = \frac{e^{2x}}{8}\left[8x^3\right] = x^3 e^{2x}$$

Since $F'(x) = x^3 e^{2x}$, the integration is verified.