Q.Calculate the area under the curve $y = 2\sqrt{x}$ included between the lines $x = 0$ and $x = 1$.

Concept: Area Under Curve — the area bounded by $y = f(x)$, the $x$-axis, and the vertical lines $x = a$, $x = b$ is given by $\int_a^b f(x) \, dx$.

Step 1: Identify the curve and limits.

$y = 2\sqrt{x}$, from $x = 0$ to $x = 1$. The curve lies above the $x$-axis in this interval.

Step 2: Set up the definite integral.

Step 3: Integrate.

Recall $\sqrt{x} = x^{1/2}$, so

Step 4: Evaluate from $0$ to $1$.

Why integration works here

When we talk about "area under a curve" between two vertical lines, we mean the region bounded by the curve $y = f(x)$, the $x$-axis, and the lines $x = a$ and $x = b$. The fundamental idea is that we slice this region into infinitely thin vertical strips of width $dx$ and height $f(x)$. The area of each strip is $f(x)\,dx$, and adding them all up gives the definite integral $\int_a^b f(x)\,dx$.

For $y = 2\sqrt{x}$, the curve lies entirely above the $x$-axis for $x \geq 0$, so no sign issues arise — the integral directly gives the geometric area.

Step-by-step calculation

1. Set up the integral.

The boundaries are $x = 0$ and $x = 1$, and the function is $y = 2\sqrt{x}$. So the area $A$ is:

2. Rewrite the integrand in power form.

Recall that $\sqrt{x} = x^{1/2}$. So:

3. Apply the power rule for integration.

For any $n \neq -1$, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$. Here $n = \frac{1}{2}$, so $n+1 = \frac{3}{2}$:

4. Evaluate the definite integral.

Using the Fundamental Theorem of Calculus:

Method: Area under a single curve between two vertical lines

Use this whenever the region is bounded above by one curve $y = f(x)$, below by the $x$-axis, and on the sides by $x = a$ and $x = b$, with $f(x) \ge 0$ throughout.

Steps

Step 1: Confirm the curve stays above the axis.

On the interval $[a, b]$, check that $f(x) \ge 0$ (for roots and even powers this is usually automatic). If it dips below, you would need $|f(x)|$ — so this check protects the sign of the answer.

Step 2: Write the area as a definite integral.

This simply sums thin vertical strips of height $f(x)$ and width $dx$.

Step 3: Rewrite roots as fractional powers, then apply the power rule.

Convert any radical, e.g. $\sqrt{x} = x^{1/2}$, and integrate term by term with

Step 4: Substitute the limits (upper minus lower).

Evaluate the antiderivative at $b$ and $a$ and subtract. A quick verification is to differentiate your antiderivative and confirm it returns the original integrand.

Common Mistakes

Mistake 1: Dropping the coefficient $2$ in $y = 2\sqrt{x}$

Why it's wrong: integrating $\sqrt{x}$ alone gives $\tfrac23 x^{3/2}$ and a final area of $\tfrac23$, but the height of every strip is $2\sqrt{x}$, not $\sqrt{x}$. Correct approach: keep the factor, $\int_0^1 2x^{1/2}\,dx = \tfrac43 x^{3/2}\big|_0^1 = \tfrac43$.

Mistake 2: Misapplying the power rule to $x^{1/2}$

Why it's wrong: students write $\int x^{1/2}\,dx = \tfrac{x^{3/2}}{3}$ or forget to raise the exponent by one. Correct approach: with $n = \tfrac12$, $\int x^{1/2}\,dx = \dfrac{x^{3/2}}{3/2} = \tfrac23 x^{3/2}$.