Area Under the Curve: Short Questions

This lesson works through two short exam questions on finding the area of a shaded region using definite integration. In the first we use a line minus a curve; in the second we integrate a single curve over a given interval.

Question 1: area between the parabola and the line

We are given the curve $x^2 = y$ (a parabola opening upward) and the line $y = 4$. The shaded region lies between the line and the curve in the first quadrant.

Finding the intersection point

Substitute $y = 4$ into the curve $x^2 = y$:

$x^2 = 4 \implies x = \pm 2.$

The shaded region is in the first quadrant, so we take $x = 2$. The curve and line meet at $(2, 4)$, and the limits on the $x$ axis run from $0$ to $2$.

Setting up the area

The shaded area is the area under the line minus the area under the curve:

$A = \int_0^2 4\,dx - \int_0^2 x^2\,dx.$

Evaluating

$A = \big[\,4x\,\big]_0^2 - \left[\tfrac{x^3}{3}\right]_0^2 = 4(2) - \tfrac{1}{3}(2^3) = 8 - \tfrac{8}{3}.$

$A = \tfrac{24 - 8}{3} = \tfrac{16}{3} \text{ square units}.$

Question 2: area enclosed by $y^2 = x$, $x = 3$ and the $x$ axis

Here no figure is given, so we sketch it ourselves. The curve $y^2 = x$ is a parabola opening to the right. With the line $x = 3$ and the $x$ axis, the shaded region sits in the first quadrant.

Setting up the area

From the curve, $y = \sqrt{x}$. Taking $x$ as the variable, the area runs from $0$ to $3$:

$A = \int_0^3 y\,dx = \int_0^3 \sqrt{x}\,dx = \int_0^3 x^{1/2}\,dx.$

Evaluating

Using $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$ with $n = \tfrac{1}{2}$:

$A = \left[\frac{x^{3/2}}{3/2}\right]_0^3 = \left[\tfrac{2}{3}\,x^{3/2}\right]_0^3 = \tfrac{2}{3}\,(3)^{3/2}.$

Since $3^{3/2} = \sqrt{3^3} = \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$:

$A = \tfrac{2}{3}\cdot 3\sqrt{3} = 2\sqrt{3} \text{ square units}.$

Key takeaways

• To find an area between a line and a curve, subtract the integral of the curve from the integral of the line over the same limits.
• Find the limits of integration from where the boundaries intersect; keep only the values that lie in the required region.
• For a sideways parabola $y^2 = x$, write $y = \sqrt{x}$ and integrate with respect to $x$, using $\int x^{1/2}\,dx = \tfrac{2}{3}x^{3/2}$.