Q.Assertion (A) : The area of the region bounded by the line $y - 1 = x$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ is 2 square units. Reason (R) : The area of the region bounded by the curve $y = f(x)$, the $x$-axis and the ordinates $x = a$ and $x = b$ is given by $\int_{a}^{b} f(x)\, dx$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

Polynomial Integration — From Intuition to Precision

Think of integration as the reverse of differentiation. If differentiation tells you the slope of a curve at every point, integration tells you the area under that curve between two points. For polynomials, this reverse process is beautifully simple.

The Intuition: Undoing the Power Rule

You already know the power rule for differentiation: if $f(x) = x^n$, then $f'(x) = n x^{n-1}$. Integration asks: given that the derivative is $x^n$, what was the original function?

Suppose you want a function whose derivative is $x^2$. You need something that, when differentiated, gives $x^2$. Try $x^3$: its derivative is $3x^2$, which is three times too big. So try $\frac{1}{3}x^3$ — its derivative is exactly $x^2$. That's the core idea: increase the exponent by 1, then divide by the new exponent.

The $+C$ is crucial. Why? Because the derivative of any constant is zero. If $F(x) = \frac{1}{3}x^3 + 5$, its derivative is still $x^2$. So when we integrate, we must add an arbitrary constant $C$ to account for all possible original functions.

The Precise Statement

For a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, its indefinite integral (antiderivative) is:

You integrate term by term, applying the reverse power rule to each term separately. The constant term $a_0$ integrates to $a_0 x$, since $\int a_0 \, dx = a_0 x$ (because the derivative of $a_0 x$ is $a_0$).

A Worked Example

Integrate $f(x) = 4x^3 - 2x + 7$.

Apply the rule term by term:

• $4x^3$: increase exponent to 4, divide by 4 → $\frac{4}{4}x^4 = x^4$
• $-2x$: this is $-2x^1$, increase exponent to 2, divide by 2 → $\frac{-2}{2}x^2 = -x^2$
• $7$: this is $7x^0$, increase exponent to 1, divide by 1 → $7x$

So:

You can check by differentiating: the derivative of $x^4 - x^2 + 7x + C$ is $4x^3 - 2x + 7$, which is exactly your original function.

Definite Integration: Area Under the Curve

When you want the actual area between $x = a$ and $x = b$, you use the definite integral: …