Q.Find the area under the given curves and given lines:

Each area is the definite integral of the curve between the given vertical lines (both curves stay above the $x$-axis on the intervals).

(i) $y=x^2$ from $x=1$ to $x=2$:

(ii) $y=x^4$ from $x=1$ to $x=5$:

The area bounded by a curve $y=f(x)$, the $x$-axis, and two vertical lines $x=a$, $x=b$ (with $f(x)\ge 0$) is the definite integral $\int_a^b f(x)\,dx$ — the sum of thin vertical strips of height $f(x)$ and width $dx$. Both curves here are positive on their intervals, so the integral gives the area directly.

(i) $y=x^2$, from $x=1$ to $x=2$

Apply the power rule $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}$ with $n=2$:

(ii) $y=x^4$, from $x=1$ to $x=5$

With $n=4$:

Method: Area under a power curve $y = x^n$ above the x-axis

Use this for the area bounded by a power curve $y = x^n$, the $x$-axis, and two vertical lines, when the curve stays above the axis on the interval.

Steps

Step 1: Confirm positivity on $[a,b]$.

For $x > 0$ and any $n$, $x^n > 0$, so the integral gives the area directly with no splitting.

Step 2: Apply the power rule.

Step 3: Evaluate between the limits, one part at a time.

When several curves are asked in one question, treat each as a separate independent integral with its own $n$ and its own limits.

Common Mistakes

Mistake 1: Adding one to the base instead of the exponent in the power rule.

Why it's wrong: $\int x^4\,dx = \tfrac{x^5}{5}$, not $\tfrac{x^5}{4}$ or $\tfrac{x^4}{5}$ — the exponent increases by one and you divide by the new exponent. Correct approach: $\int x^n dx = \tfrac{x^{n+1}}{n+1}$, so part (ii) is $[\tfrac{x^5}{5}]_1^5 = \tfrac{3124}{5}$.

Mistake 2: Arithmetic slips with large powers.

Why it's wrong: $5^5 = 3125$ (not $625$), and forgetting to subtract the lower-limit term $\tfrac{1}{5}$ changes the answer. Correct approach: compute $\tfrac{3125-1}{5} = \tfrac{3124}{5}$; and for part (i), $\tfrac{8-1}{3}=\tfrac73$.