Let $f(x)$ be a continuous function on $[0,1]$ such that $f(1)=0$. $\int_0^1 (f'(x))^2.dx=7$ and $\int_0^1 x^2f(x).dx=\frac13$. Find $\int_0^1f(x).dx$

Question

I have a solution for the above question but i wanted to check if what i am doing is correct and can be done or not.

The function i am getting also satisfies both the condition but i still am not sure if my method is correct or not.

I have given my solution below in the answers

, is an image of the original question

Answer 1

According to the first crucial step of the solution proposed by the asker we get $$\int\limits_0^1x^3f'(x)\,dx =-1\quad (*)$$ By the Cauchy-Schwarz inequality we obtain $$1\le \int\limits_0^1x^6\,dx\, \int\limits_0^1[f'(x)]^2\,dx={1\over 7}\cdot 7=1$$ As the equality occurs in the C-S inequality, we get $f'(x)=cx^3.$ Next $(*)$ gives $c=-7.$ As $f(1)=0$ we get $$f(x)=-{7\over 4}(x^3-1)$$

Remark The solution can be made more elementary, by making a guess that $f'(x)=cx^3.$ Then $(*)$ gives $c=-7.$ Then we observe that $$\int\limits_0^1\left [f'(x)+7x^3\right ]^2\,dx=0$$ so $f'(x)=-7x^3.$

Answer 2

Its given: $\int_0^1 x^2f(x).dx = \frac13$

Applying Integration By Parts, and using the condition $f(1)=0$ we arrive at the situation,

$\int_0^1 x^3f'(x).dx = -1$

Now let, $(f'(x)-cx^3)^2=0$

$\implies \int_0^1(f'(x)-cx^3)^2.dx=0$

$\implies \int_0^1 (f'(x))^2.dx + c^2\int_0^1x^6.dx -2c\int_0^1x^3f'(x).dx = 0$

putting values,

$\implies 7 + \frac{c^2}7 +2c= 0$

multiplying whole eqn by 7 gives,

$\implies c^2+49+14c=(c+7)^2=0 \implies c=-7$

this gives,

$(f'(x)+7x^3)^2=0 \implies f'(x)=-7x^3$

Integrating we get,

$f(x)=\frac{-7x^4}4+C$, where C=constant of integration

using $f(1)=0\implies C=\frac 74$

$f(x)=\frac{-7x^4}4 + \frac 74$

Now thus arriving finally at the answer by definite integration of $f(x)$ as,

$\int_0^1f(x).dx=\frac 75$

Answer 3

$f(x)$ is a continuous function on $[0,1].f(1) = 0$.

$\int_0^1 (f’(x))^2 , dx = 7$

$\int_0^1 x^2 f(x).dx = \frac13$.

We want to find $\int_0^1 f(x).dx$.

First, let’s use the given information to find an expression for $f’(x)$:

We know that $\int_0^1 (f’(x))^2.dx = 7$.

Applying the fundamental theorem of calculus, we have:

$\int_0^1 (f’(x))^2.dx = f(1) - f(0) = 0 - f(0) = -f(0)$.

Therefore, $-f(0) = 7$, which implies $f(0) = -7$

Next, let’s find an expression for $\int_0^1 x^2 f(x).dx$:

$\int_0^1 x^2 f(x).dx=\frac13$

Now, let’s use integration by parts to relate $\int_0^1 f(x).dx$ to the given integral: $\int_0^1 x^2 f(x).dx = \frac13$. Let $u = x^2$ and $dv = f(x).dx$.

Then, we have: $du = 2x.dx$ and $v = \int f(x).dx$

Applying integration by parts: $uv - \int v.du = \frac13$

$x^2 f(x) - \int 2x f(x).dx = \frac13$

$x^2 f(x) - 2 \int x f(x).dx = \frac13$

$x^2 f(x) - 2 \left(\int_0^1 x f(x).dx\right) = \frac13$

Rearranging: $\int_0^1 x f(x).dx = \frac{x^2 f(x) - \frac13}{2}$

Now, let’s find $\int_0^1 f(x).dx:$

$\int_0^1 f(x).dx = f(1) - f(0) = 0 - (-7) = 7$

Therefore, the value of $\int_0^1 f(x).dx$ is 7.