Definite Integral of Cubic Term and Second Derivative

To evaluate the given definite integral $I = \int_{0}^{1} 4x^3 \left\{ \frac{d^2}{dx^2}(1 - x^2)^5 \right\} dx$

Let $f(x) = (1 - x^2)^5$. The integral can be written as: $I = \int_{0}^{1} 4x^3 f''(x) \, dx$

We apply integration by parts with $u = 4x^3$ and $dv = f''(x) \, dx$:

• $du = 12x^2 \, dx$
• $v = f'(x)$

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$: $I = \left[ 4x^3 f'(x) \right]_{0}^{1} - \int_{0}^{1} 12x^2 f'(x) \, dx$

First, let's find $f'(x)$: $f'(x) = \frac{d}{dx} (1 - x^2)^5 = 5(1 - x^2)^4 \cdot (-2x) = -10x(1 - x^2)^4$

Evaluating the boundary term $\left[ 4x^3 f'(x) \right]_{0}^{1}$:

• At $x = 1$, $f'(1) = -10(1)(1 - 1^2)^4 = 0$, so $4(1)^3 f'(1) = 0$.
• At $x = 0$, $4(0)^3 f'(0) = 0$.

Thus, the boundary term vanishes, and we get: $I = -\int_{0}^{1} 12x^2 f'(x) \, dx$

Now, we apply integration by parts a second time with $u = 12x^2$ and $dv = f'(x) \, dx$:

• $du = 24x \, dx$
• $v = f(x)$

$I = -\left( \left[ 12x^2 f(x) \right]_{0}^{1} - \int_{0}^{1} 24x f(x) \, dx \right)$

Evaluating the boundary term $\left[ 12x^2 f(x) \right]_{0}^{1}$:

• At $x = 1$, $f(1) = (1 - 1^2)^5 = 0$, so $12(1)^2 f(1) = 0$.
• At $x = 0$, $12(0)^2 f(0) = 0$.

This boundary term also vanishes, simplifying the integral to: $I = \int_{0}^{1} 24x f(x) \, dx = \int_{0}^{1} 24x (1 - x^2)^5 \, dx$

To compute this final integral, we use the substitution $t = 1 - x^2$:

• $dt = -2x \, dx \implies x \, dx = -\frac{1}{2} dt$
• When $x = 0$, $t = 1$
• When $x = 1$, $t = 0$

Substituting these into the integral gives: $I = 24 \int_{1}^{0} t^5 \left( -\frac{1}{2} \right) dt = 12 \int_{0}^{1} t^5 \, dt$

Evaluating this simple polynomial integral: $I = 12 \left[ \frac{t^6}{6} \right]_{0}^{1} = 12 \left( \frac{1}{6} - 0 \right) = 2$