Mathematics - Definite Integration Question with Solution | TestHub

number of solution is zero
The current solution "number of solution is zero" is incorrect and does not address the question of computing $(FG{)}^{\prime }(0)$. The correct answer is 0.

Here's a detailed explanation:

Problem:

Let $F(x)={\int }_{-1}^x\sqrt{4+t^2}dt$ and $G(x)={\int }_x^1\sqrt{4+t^2}dt$. Compute the value of $(FG{)}^{\prime }(0)$.

Explanation:

1. Understand the functions $F(x)$ and $G(x)$:

$F(x)$ is an integral with a variable upper limit.

$G(x)$ is an integral with a variable lower limit.

2. Apply the Fundamental Theorem of Calculus (Part 1) to find $F^{\prime }(x)$ and $G^{\prime }(x)$:

The Fundamental Theorem of Calculus states that if $H(x)={\int }_a^{x}f(t)dt$, then $H^{\prime }(x)=f(x)$.

For $F(x)={\int }_{-1}^x\sqrt{4+t^2}dt$:

$F^{\prime }(x)=\sqrt{4+x^2}$

For $G(x)={\int }_x^1\sqrt{4+t^2}dt$:

We can rewrite $G(x)$ as $G(x)=-{\int }_1^x\sqrt{4+t^2}dt$.

Therefore, $G^{\prime }(x)=-\sqrt{4+x^2}$.

3. Apply the Product Rule for Differentiation:

The product rule states that $(FG{)}^{\prime }(x)=F^{\prime }(x)G(x)+F(x)G^{\prime }(x)$.

We need to compute $(FG{)}^{\prime }(0)$, so we substitute $x=0$:

$(FG{)}^{\prime }(0)=F^{\prime }(0)G(0)+F(0)G^{\prime }(0)$.

4. Calculate $F(0)$, $G(0)$, $F^{\prime }(0)$, and $G^{\prime }(0)$:

Calculate $F(0)$:

$F(0)={\int }_{-1}^0\sqrt{4+t^2}dt$

Calculate $G(0)$:

$G(0)={\int }_0^1\sqrt{4+t^2}dt$

Calculate $F^{\prime }(0)$:

$F^{\prime }(0)=\sqrt{4+0^2}=\sqrt{4}=2$

Calculate $G^{\prime }(0)$:

$G^{\prime }(0)=-\sqrt{4+0^2}=-\sqrt{4}=-2$

5. Substitute these values into the product rule formula:

$(FG{)}^{\prime }(0)=(2)\left({\int }_0^1\sqrt{4+t^2}dt\right)+\left({\int }_{-1}^0\sqrt{4+t^2}dt\right)(-2)$

$(FG{)}^{\prime }(0)=2{\int }_0^1\sqrt{4+t^2}dt-2{\int }_{-1}^0\sqrt{4+t^2}dt$

6. Notice the relationship between $F(x)$ and $G(x)$:

$F(x)+G(x)={\int }_{-1}^x\sqrt{4+t^2}dt+{\int }_x^1\sqrt{4+t^2}dt$

Using the property of definite integrals ${\int }_a^{b}f(t)dt+{\int }_b^{c}f(t)dt={\int }_a^{c}f(t)dt$:

$F(x)+G(x)={\int }_{-1}^1\sqrt{4+t^2}dt$.

This means $F(x)+G(x)$ is a constant with respect to $x$.

Let $C={\int }_{-1}^1\sqrt{4+t^2}dt$.

So, $F(x)+G(x)=C$.

Differentiating both sides with respect to $x$:

$F^{\prime }(x)+G^{\prime }(x)=0$.

This implies $G^{\prime }(x)=-F^{\prime }(x)$.

We already found $F^{\prime }(x)=\sqrt{4+x^2}$ and $G^{\prime }(x)=-\sqrt{4+x^2}$, which confirms this relationship.

7. Let's re-evaluate $F(0)$ and $G(0)$ using the relationship $F(x)+G(x)=C$:

$F(0)={\int }_{-1}^0\sqrt{4+t^2}dt$

$G(0)={\int }_0^1\sqrt{4+t^2}dt$

Consider the integrand $f(t)=\sqrt{4+t^2}$. This is an even function because $f(-t)=\sqrt{4+(-t{)}^2}=\sqrt{4+t^2}=f(t)$.

For an even function, ${\int }_{-a}^{a}f(t)dt=2{\int }_0^{a}f(t)dt$.

Also, ${\int }_{-a}^{0}f(t)dt={\int }_0^{a}f(t)dt$.

Therefore, $F(0)={\int }_{-1}^0\sqrt{4+t^2}dt={\int }_0^1\sqrt{4+t^2}dt=G(0)$.

Let $K={\int }_0^1\sqrt{4+t^2}dt$.

So, $F(0)=K$ and $G(0)=K$.

8. Substitute these values back into the product rule formula:

$(FG{)}^{\prime }(0)=F^{\prime }(0)G(0)+F(0)G^{\prime }(0)$

$(FG{)}^{\prime }(0)=(2)(K)+(K)(-2)$

$(FG{)}^{\prime }(0)=2K-2K$

$(FG{)}^{\prime }(0)=0$

The final answer is $\boxed{0}$.