Mathematics - Definite Integration Question with Solution | TestHub

We have $F(x)+F\left(x+\frac{1}{2}\right)=3\dots \dots (1)$.

Replace $x$ by $x+\frac{1}{2}$ in (1), we get $F\left(x+\frac{1}{2}\right)+F(x+1)=3\dots \dots (2)$.

From (1) and (2), we get $F(x)=F(x+1)\dots \dots (3)$, which implies $F(x)$ is a periodic function.

Now consider:

$$I={\int }_0^{1500}F(x)dx=1500{\int }_0^{1}F(x)dx$$

Using the property of periodic functions:

$$I=1500\left[{\int }_0^{\frac{1}{2}}F(x)dx+{\int }_{\frac{1}{2}}^{1}F(x)dx\right]$$

In the second integral, let $y=x-\frac{1}{2}$, so $x=y+\frac{1}{2}$. When $x=\frac{1}{2}$, $y=0$. When $x=1$, $y=\frac{1}{2}$.

$$I=1500\left[{\int }_0^{\frac{1}{2}}F(x)dx+{\int }_0^{\frac{1}{2}}F\left(y+\frac{1}{2}\right)dy\right]$$

$$I=1500{\int }_0^{\frac{1}{2}}\left(F(x)+F\left(x+\frac{1}{2}\right)\right)dx$$

Using (1), we have $F(x)+F\left(x+\frac{1}{2}\right)=3$.

$$I=1500{\int }_0^{\frac{1}{2}}3dx$$

Hence,

$$I=1500(3)\left(\frac{1}{2}\right)=750\times 3=2250$$