he area (in sq. units) of the region bounded by the parabola y² = 4x and the line x = 1 is

The correct answer is:

$\frac{8}{3}$

Step-by-Step Explanation:

To find the area of the region bounded by the parabola $y^2=4x$ and the line $x=1$, we can integrate with respect to $x$.

The equation of the parabola is given by:

$y^2=4x$

Taking the square root on both sides, we get the upper and lower halves of the parabola:

$y=\pm 2\sqrt{x}$

The region is bounded between $x=0$ (the vertex of the parabola) and the vertical line $x=1$. Since the parabola is symmetric about the x-axis, the total area is twice the area of the upper region (from $y=0$ to $y=2\sqrt{x}$).

Therefore, the area $A$ of the bounded region can be set up as the definite integral:

$A=2{\int }_0^{1}2\sqrt{x}dx$

Simplifying the integral, we pull out the constant factor $2$:

$A=4{\int }_0^1{x}^{\frac{1}{2}}dx$

Now, we integrate $x^{1/2}$ using the power rule of integration, which states that $\int x^{n}dx=\frac{x^{n+1}}{n+1}$:

$A=4{\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^1$

Simplifying the exponent and the denominator:

$A=4{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1$

This simplifies to:

$A=4·\frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^1$

Evaluating at the upper limit ($x=1$) and lower limit ($x=0$):

$A=\frac{8}{3}\left(1^{\frac{3}{2}}-0^{\frac{3}{2}}\right)$

Thus, we compute the final bounded area:

$A=\frac{8}{3}·1=\frac{8}{3}$

Therefore, the area of the region bounded by the parabola and the line is $\frac{8}{3}$ square units.