Question

Let g be a differentiable function satisfying

$\int_{0}^{x}(x-t+1)g(t)dt=x^4+x^2 \forall x \geq 0$, then the value of

$\int_{0}^{1}\frac{12}{g'(x)+g(x)+10}dx=$____.

Answer 1

$\frac{\pi}{4}$

Answer 2

The given integral equation is:

$\int_{0}^{x}(x-t+1)g(t)dt=x^4+x^2 \quad \forall x \geq 0$

We can split the integral on the left-hand side:

$\int_{0}^{x}(x+1)g(t)dt - \int_{0}^{x}tg(t)dt = x^4+x^2$

Since $(x+1)$ is independent of $t$, we can take it out of the first integral:

$(x+1)\int_{0}^{x}g(t)dt - \int_{0}^{x}tg(t)dt = x^4+x^2 \quad (*)$

Now, we differentiate both sides with respect to $x$. We use the Leibniz integral rule, which states that if $F(x) = \int_{a(x)}^{b(x)} f(x,t) dt$, then $F'(x) = f(x, b(x)) b'(x) - f(x, a(x)) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) dt$.

For the first term, $(x+1)\int_{0}^{x}g(t)dt$:

Using the product rule and the Fundamental Theorem of Calculus ($\frac{d}{dx}\int_0^x g(t)dt = g(x)$):

$\frac{d}{dx}\left[ (x+1)\int_{0}^{x}g(t)dt \right] = 1 \cdot \int_{0}^{x}g(t)dt + (x+1) \cdot g(x)$

$= \int_{0}^{x}g(t)dt + xg(x) + g(x)$

For the second term, $\int_{0}^{x}tg(t)dt$:

Using the Fundamental Theorem of Calculus:

$\frac{d}{dx}\left[ \int_{0}^{x}tg(t)dt \right] = xg(x)$

Differentiating the right-hand side, $x^4+x^2$:

$\frac{d}{dx}(x^4+x^2) = 4x^3+2x$

Now, substitute these differentiated terms back into the differentiated equation $(*)$:

$\left( \int_{0}^{x}g(t)dt + xg(x) + g(x) \right) - xg(x) = 4x^3+2x$

$\int_{0}^{x}g(t)dt + g(x) = 4x^3+2x \quad (1)$

We need to find $g'(x)+g(x)$, so we differentiate equation (1) with respect to $x$ again.

Differentiating $\int_{0}^{x}g(t)dt$ gives $g(x)$ (by Fundamental Theorem of Calculus).

Differentiating $g(x)$ gives $g'(x)$.

Differentiating $4x^3+2x$ gives $12x^2+2$.

So, differentiating equation (1) yields:

$g(x) + g'(x) = 12x^2+2$

Now we need to evaluate the integral $\int_{0}^{1}\frac{12}{g'(x)+g(x)+10}dx$.

Substitute the expression for $g'(x)+g(x)$:

$\int_{0}^{1}\frac{12}{(12x^2+2)+10}dx$

$= \int_{0}^{1}\frac{12}{12x^2+12}dx$

Factor out 12 from the denominator:

$= \int_{0}^{1}\frac{12}{12(x^2+1)}dx$

$= \int_{0}^{1}\frac{1}{x^2+1}dx$

This is a standard integral, $\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)$. Here $a=1$.

$= \left[ \arctan(x) \right]_{0}^{1}$

$= \arctan(1) - \arctan(0)$

$= \frac{\pi}{4} - 0$

$= \frac{\pi}{4}$