Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12
Question:
If $G(x,t) = \begin{cases} x(t-1), & \text{when } x \leq t \\ t(x-1), & \text{when } t < x \end{cases}$ and if $f$ is continuous function of $x$ in $[0,1]$. Let $g(x) = \int_0^1 f(t)G(x,t)dt$. Then which is incorrect:
$g(0) + g(1) = 0$
$g(0) = 0$
$g(1) = 1$
$g''(x) = f(x)$
Step-by-Step Solution
Key Concept: Double differentiation of a convolution-type integral recovers the original integrand function.
For $t \geq 0$, $G(0,t) = 0$; for $t < 1$, $G(1,t) = t(1-1) = 0$ so $g(1) = \int_0^1 t(1)·0\,dt = 0$. Using the product rule: $g(x) = \int_0^x f(t)(x-1)dt + \int_x^1 f(t)(t-1)dt = (x-1)\int_0^x uf(u)du + x\int_x^1 f(t)(t-1)dt$. Differentiating yields $g'(x) = xf(x) + \int_0^x f(t)dt + \int_x^1 f(t)(t-1)dt - xf(x)(x-1) = \int_0^x f(t)dt - \int_0^x f(t)dt + f(x)$, so $g''(x) = f(x)$.
Correct Answer: 3