Mathematics - Area Under Curves Question with Solution | TestHub

The equation of the curve is given by $y=a\sqrt{x}+bx$.

Since the curve passes through the point $(1,2)$, we substitute $x=1$ and $y=2$ into the equation:

$a\sqrt{1}+b(1)=2$

$a+b=2$

The area bounded by the curve, the line $x=4$, and the X-axis is $8$ sq. units. This can be written as:

${\int }_0^4(a\sqrt{x}+bx)dx=8$

${\left[a\frac{x^{3/2}}{3/2}+b\frac{x^2}{2}\right]}_0^4=8$

${\left[\frac{2a}{3}{x}^{3/2}+\frac{b}{2}{x}^2\right]}_0^4=8$

Substituting the upper limit $x=4$:

$\frac{2a}{3}(4{)}^{3/2}+\frac{b}{2}(4{)}^2=8$

$\frac{2a}{3}(8)+\frac{b}{2}(16)=8$

$\frac{16a}{3}+8b=8$

Dividing the entire equation by $8$:

$\frac{2a}{3}+b=1$

$2a+3b=3$

Now, we solve the two linear equations:
$a+b=2$
$2a+3b=3$

From the first equation, $b=2-a$. Substituting this into the second equation:

$2a+3(2-a)=3$

$2a+6-3a=3$

$-a=-3\Rightarrow a=3$

Substituting $a=3$ back into $b=2-a$:

$b=2-3=-1$

We need to find the value of $a-b$:

$a-b=3-(-1)=4$

Answer: $4$