Q.Draw a rough sketch of the curve $y = \sqrt{x - 1}$ in the interval $[1, 5]$. Find the area under the curve and between the lines $x = 1$ and $x = 5$.

Concept: Area Under Curve — the definite integral of $y$ with respect to $x$ over the given interval.

Step 1 – Sketch the curve

$y = \sqrt{x - 1}$ is defined for $x \ge 1$. It starts at $(1, 0)$ and increases slowly. At $x = 5$, $y = \sqrt{4} = 2$. The shape is the upper half of a rightward-opening parabola.

Step 2 – Set up the area integral

Area $= \int_{1}^{5} \sqrt{x - 1} \, dx$.

Step 3 – Evaluate

Let $u = x - 1$, $du = dx$. When $x = 1$, $u = 0$; when $x = 5$, $u = 4$. …

The problem asks for the area under the curve $y = \sqrt{x-1}$ between $x=1$ and $x=5$. This is a straightforward application of definite integration — the area bounded by the curve, the x-axis, and the vertical lines $x=1$ and $x=5$.

Why integration works here: For a non-negative function $y = f(x)$ on $[a, b]$, the area between the curve and the x-axis is exactly $\int_a^b f(x)\,dx$. Each tiny vertical strip of width $dx$ has height $f(x)$, so its area is $f(x)\,dx$. Summing (integrating) these strips gives the total area.

Let's work through it.

1. Sketch the curve

   $y = \sqrt{x-1}$ is defined for $x \ge 1$. It's the upper half of a rightward-opening parabola with vertex at $(1,0)$. At $x=1$, $y=0$; at $x=5$, $y = \sqrt{4} = 2$. The curve rises smoothly, concave downward (since the second derivative is negative). The region is a simple curved shape sitting above the x-axis from $x=1$ to $x=5$.

2. Set up the integral

   The area $A$ is given by:

3. Substitute to simplify Let $u = x-1$. Then $du = dx$, and when $x=1$, $u=0$; when $x=5$, $u=4$. The integral becomes:

4. Integrate Using the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$:

5. Evaluate …

Method: Area under a shifted-root curve

Use this for areas under curves such as $y = \sqrt{x - h}$ — a basic square-root graph translated sideways — between two vertical lines.

Steps

Step 1: Identify the domain and starting point.

$y = \sqrt{x - h}$ exists only for $x \ge h$ and starts at the point $(h, 0)$, rising slowly. Sketching this tells you the curve is above the $x$-axis throughout the interval, so no sign correction is needed.

Step 2: Set up the area integral.

Step 3: Substitute to remove the shift. …

Common Mistakes

Mistake 1: Substituting $u = x - 1$ but keeping the old limits $1$ and $5$

Why it's wrong: once you switch the variable to $u$, the limits must switch too ($x = 1 \to u = 0$, $x = 5 \to u = 4$); evaluating $\tfrac23 u^{3/2}$ at $u = 1$ and $u = 5$ gives a wrong number. Correct approach: either change the limits to $0$ and $4$, or back-substitute to $\tfrac23(x-1)^{3/2}$ before using $1$ and $5$. …