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Evaluate the integral. int2x^(2)4^(-2x)dx Answer:

Evaluate the integral.\newline∫2x24−2xdx \int 2 x^{2} 4^{-2 x} d x \newlineAnswer:

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Find indefinite integrals using the substitution and by partsright chevron icon

Full solution

Q. Evaluate the integral.\newline∫2x24−2xdx \int 2 x^{2} 4^{-2 x} d x \newlineAnswer:
  1. Denote Integral: Let's denote the integral by II:I=∫2x2⋅4−2x dxI = \int 2x^2 \cdot 4^{-2x} \, dxTo solve this integral, we will use integration by parts, which states that ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du, where uu and dvdv are differentiable functions of xx. We need to choose uu and dvdv such that the resulting integral is simpler to solve. Let's choose:u=x2u = x^2 (which implies du=2x dxdu = 2x \, dx)dv=2⋅4−2x dxdv = 2 \cdot 4^{-2x} \, dx (which implies v=∫2⋅4−2x dxv = \int 2 \cdot 4^{-2x} \, dx) Now we need to find vv by integrating dvdv.
  2. Integration by Parts: To find vv, we integrate dvdv:
    v=∫2⋅4−2x dxv = \int 2 \cdot 4^{-2x} \, dx
    We can rewrite 4−2x4^{-2x} as (22)−2x=2−4x(2^2)^{-2x} = 2^{-4x}.
    Now, v=∫2⋅2−4x dxv = \int 2 \cdot 2^{-4x} \, dx
    To integrate this, we use the fact that ∫akx dx=(1k⋅ln⁡(a))⋅akx\int a^{kx} \, dx = (\frac{1}{k} \cdot \ln(a)) \cdot a^{kx}, where a > 0, a≠1a \neq 1, and kk is a constant.
    dvdv00
    dvdv11
    Now we have vv, we can proceed to apply the integration by parts formula.
  3. Find vv: Applying integration by parts:\newlineI=uv−∫vduI = uv - \int v du\newlineI=x2∗(−1/(2ln⁡(2))∗2−4x)−∫(−1/(2ln⁡(2))∗2−4x)∗2xdxI = x^2 * (-1/(2\ln(2)) * 2^{-4x}) - \int(-1/(2\ln(2)) * 2^{-4x}) * 2x dx\newlineSimplify the expression:\newlineI=−x22ln⁡(2)∗2−4x+1ln⁡(2)∫x∗2−4xdxI = -\frac{x^2}{2\ln(2)} * 2^{-4x} + \frac{1}{\ln(2)} \int x * 2^{-4x} dx\newlineNow we need to integrate the remaining integral, which again requires integration by parts.
  4. Apply Integration by Parts: We need to integrate ∫x⋅2−4x dx\int x \cdot 2^{-4x} \, dx by parts again. Let's choose new uu and dvdv:
    u=xu = x (which implies du=dxdu = dx)
    dv=2−4x dxdv = 2^{-4x} \, dx (which implies v=∫2−4x dxv = \int 2^{-4x} \, dx)
    We already found the integral of 2−4x dx2^{-4x} \, dx in a previous step, so:
    v=−14ln⁡(2)⋅2−4xv = -\frac{1}{4\ln(2)} \cdot 2^{-4x}
    Now we apply integration by parts again.
  5. Integrate Again: Applying integration by parts to the new integral:\newline∫x⋅2−4x dx=uv−∫v du\int x \cdot 2^{-4x} \, dx = uv - \int v \, du\newline= x⋅(−12ln⁡(2)⋅2−4x)−∫(−12ln⁡(2)⋅2−4x) dxx \cdot \left(-\frac{1}{2\ln(2)} \cdot 2^{-4x}\right) - \int\left(-\frac{1}{2\ln(2)} \cdot 2^{-4x}\right) \, dx\newlineSimplify the expression:\newline= −x2ln⁡(2)⋅2−4x+12ln⁡(2)∫2−4x dx-\frac{x}{2\ln(2)} \cdot 2^{-4x} + \frac{1}{2\ln(2)} \int 2^{-4x} \, dx\newlineWe already know the integral of 2−4x dx2^{-4x} \, dx, so we can write it down directly:\newline= −x2ln⁡(2)⋅2−4x+12ln⁡(2)⋅(−14ln⁡(2))⋅2−4x-\frac{x}{2\ln(2)} \cdot 2^{-4x} + \frac{1}{2\ln(2)} \cdot \left(-\frac{1}{4\ln(2)}\right) \cdot 2^{-4x}\newline= −x2ln⁡(2)⋅2−4x−18(ln⁡(2))2⋅2−4x-\frac{x}{2\ln(2)} \cdot 2^{-4x} - \frac{1}{8(\ln(2))^2} \cdot 2^{-4x}\newlineNow we can substitute this result back into our original integration by parts formula.
  6. Apply Integration Again: Substituting the result back into the original integration by parts formula:\newlineI=−x22ln⁡(2)⋅2−4x+1ln⁡(2)⋅(−x2ln⁡(2)⋅2−4x−18(ln⁡(2))2⋅2−4x)I = -\frac{x^2}{2\ln(2)} \cdot 2^{-4x} + \frac{1}{\ln(2)} \cdot \left(-\frac{x}{2\ln(2)} \cdot 2^{-4x} - \frac{1}{8(\ln(2))^2} \cdot 2^{-4x}\right)\newlineCombine like terms:\newlineI=−x22ln⁡(2)⋅2−4x−x2(ln⁡(2))2⋅2−4x−18(ln⁡(2))3⋅2−4x+CI = -\frac{x^2}{2\ln(2)} \cdot 2^{-4x} - \frac{x}{2(\ln(2))^2} \cdot 2^{-4x} - \frac{1}{8(\ln(2))^3} \cdot 2^{-4x} + C\newlineThis is the final result of the integral, where CC is the constant of integration.

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