Question

If $f(0) = 1, f(2) = 3$ and $f'(2) = 5$, then the value of the definite integral $\int_0^1 x f''(2x) dx$ is equal to ______.

Answer 1

2

Answer 2

To evaluate the definite integral $\int_0^1 x f''(2x) dx$, we use the method of integration by parts.

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

Let's choose our $u$ and $dv$: Let $u = x$ Then $du = dx$

Let $dv = f''(2x) dx$ To find $v$, we integrate $dv$: $v = \int f''(2x) dx$ To solve this integral, let $t = 2x$. Then $dt = 2 dx$, which implies $dx = \frac{1}{2} dt$. Substituting these into the integral for $v$: $v = \int f''(t) \left(\frac{1}{2}\right) dt = \frac{1}{2} \int f''(t) dt = \frac{1}{2} f'(t) = \frac{1}{2} f'(2x)$.

Now, apply the integration by parts formula to the definite integral: $\int_0^1 x f''(2x) dx = \left[ x \cdot \frac{1}{2} f'(2x) \right]_0^1 - \int_0^1 \frac{1}{2} f'(2x) dx$

Let's evaluate the first term: $\left[ \frac{x}{2} f'(2x) \right]_0^1 = \left( \frac{1}{2} f'(2 \cdot 1) \right) - \left( \frac{0}{2} f'(2 \cdot 0) \right)$ $= \frac{1}{2} f'(2) - 0$ We are given $f'(2) = 5$. So, the first term is $\frac{1}{2} \cdot 5 = \frac{5}{2}$.

Now, let's evaluate the second integral: $\int_0^1 \frac{1}{2} f'(2x) dx$. Again, use the substitution $t = 2x$, so $dt = 2 dx$, which means $dx = \frac{1}{2} dt$. The limits of integration also change: When $x=0$, $t = 2(0) = 0$. When $x=1$, $t = 2(1) = 2$. So, the integral becomes: $\int_0^2 \frac{1}{2} f'(t) \left(\frac{1}{2}\right) dt = \frac{1}{4} \int_0^2 f'(t) dt$ We know that $\int f'(t) dt = f(t)$. So, $\frac{1}{4} \int_0^2 f'(t) dt = \frac{1}{4} [f(t)]_0^2 = \frac{1}{4} (f(2) - f(0))$. We are given $f(2) = 3$ and $f(0) = 1$. So, the second term is $\frac{1}{4} (3 - 1) = \frac{1}{4} (2) = \frac{1}{2}$.

Finally, combine the results from the two parts: $\int_0^1 x f''(2x) dx = \text{First Term} - \text{Second Term}$ $\int_0^1 x f''(2x) dx = \frac{5}{2} - \frac{1}{2}$ $\int_0^1 x f''(2x) dx = \frac{4}{2} = 2$.

The final answer is $\boxed{2}$.

Explanation of the solution: The integral $\int_0^1 x f''(2x) dx$ is solved using integration by parts.

1. Set $u=x$ and $dv=f''(2x)dx$.
2. Calculate $du=dx$ and $v=\frac{1}{2}f'(2x)$.
3. Apply the integration by parts formula: $\left[ \frac{x}{2} f'(2x) \right]_0^1 - \int_0^1 \frac{1}{2} f'(2x) dx$.
4. Evaluate the first term using $f'(2)=5$: $\frac{1}{2}f'(2) - 0 = \frac{5}{2}$.
5. Evaluate the second integral using substitution $t=2x$: $\frac{1}{4} \int_0^2 f'(t) dt = \frac{1}{4} [f(t)]_0^2$.
6. Use $f(2)=3$ and $f(0)=1$ for the second term: $\frac{1}{4}(f(2)-f(0)) = \frac{1}{4}(3-1) = \frac{1}{2}$.
7. Subtract the second term from the first: $\frac{5}{2} - \frac{1}{2} = 2$.