(a) Evaluate : ∫₀² x²dx and hence show the region on the graph whose area it represents. OR (b) Evaluate : ∫₀¹

Question

\int_0^2 x^2\,dx

(OR)

(b) Evaluate :

\int_0^1 \dfrac{e^{-x}}{1+e^x}\,dx

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

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Concept understanding — Polynomial Integration

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.

Tip

The reverse power rule

To integrate $x^n$ (where $n \neq -1$ ), do:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$

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:

\int P(x) \, dx = \frac{a_n}{n+1} x^{n+1} + \frac{a_{n-1}}{n} x^n + \dots + \frac{a_1}{2} x^2 + a_0 x + C

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 $a_0$ ).

Watch out

The $n = -1$ exception

The reverse power rule $\int x^n dx = \frac{x^{n+1}}{n+1}$ fails when $n = -1$ , because you'd be dividing by zero. That case ( $\int \frac{1}{x} dx$ ) gives $\log|x| + C$ , not a power of $x$ . For polynomials, this never arises — polynomial exponents are non-negative integers.

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:

\int (4x^3 - 2x + 7) \, dx = x^4 - x^2 + 7x + C

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.

Important

The check

Differentiation is the proof of integration. Always verify your answer by differentiating it — you should recover the original integrand.

Definite Integration: Area Under the Curve

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

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$\int_0^2 x^2dx=8/3$ , the area under $y=x^2$ on $[0,2]$ .
Substituting $t=e^x$ and using partial fractions gives $\ln\!\big(\tfrac{1+e}{2e}\big)-\tfrac1e+1$ .

\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C

Part (a):

$\displaystyle\int_0^2 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^2$ .
$=\dfrac{2^3}{3}-\dfrac{0^3}{3}=\dfrac{8}{3}$ .
Region: this is the area bounded by the parabola $y=x^2$ , the $x$ -axis, and the ordinates $x=0$ and $x=2$ — the shaded region under the curve rising from the origin $(0,0)$ to the point $(2,4)$ in the first quadrant.

Part (b): $I=\displaystyle\int_0^1\dfrac{e^{-x}}{1+e^x}\,dx=\int_0^1\dfrac{dx}{e^{x}(1+e^{x})}$ .

Put $t=e^{x}\Rightarrow dt=e^{x}\,dx\Rightarrow dx=\dfrac{dt}{t}$ ; limits: $x=0\Rightarrow t=1$ , $x=1\Rightarrow t=e$ .
$I=\displaystyle\int_1^{e}\dfrac{1}{t(1+t)}\cdot\dfrac{dt}{t}=\int_1^{e}\dfrac{dt}{t^{2}(1+t)}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/W1XZY/41 markMCQ
$y - 1 = x$
›Reveal solutionSolution
$\int_{-1}^{1}(x+1)\,dx=2$ , so the Assertion is true; the Reason gives exactly the formula used, so it correctly explains it.

$=\displaystyle\int_{a}^{b} f(x)\,dx$

Rewrite the line: $y-1=x\Rightarrow y=x+1$ ; over $[-1,1]$ it is $\ge0$ (it touches the axis at $x=-1$ ).

Set up the area: $\text{Area}=\displaystyle\int_{-1}^{1}(x+1)\,dx$ .

Integrate: $\left[\dfrac{x^2}{2}+x\right]_{-1}^{1}=\left(\dfrac12+1\right)-\left(\dfrac12-1\right)=\dfrac32-\left(-\dfrac12\right)=2$ . …
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CBSE 2024Set 465/S/RQPS/41 markMCQ
$\displaystyle\int 2^{2x} \cdot 3^x \, dx$
›Reveal solutionSolution

$2^{2x}3^x = 12^x$ , and $\int a^x dx = \dfrac{a^x}{\log a}+C$ , giving $\dfrac{12^x}{\log 12}+C$ .

$\displaystyle\int a^{x}\,dx = \frac{a^{x}}{\log a} + C$
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CBSE 2023Set 465/EF1GH/41 markMCQ
$\int (x-1)e^{-x}\, dx$
(a) $(x-2)e^{-x} + C$
(b) $xe^{-x} + C$
(c) $-xe^{-x} + C$
(d) $(x+1)e^{-x} + C$
›Reveal solutionSolution
$\int (x-1)e^{-x}\,dx=-xe^{-x}+C$ , verified by differentiation.

$\int u\,dv=uv-\int v\,du$

With $u=x-1\Rightarrow du=dx$ and $dv=e^{-x}dx\Rightarrow v=-e^{-x}$ .

$\int(x-1)e^{-x}dx=(x-1)(-e^{-x})-\int(-e^{-x})dx=-(x-1)e^{-x}-e^{-x}+C$ . …
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Q.(a) Evaluate : ∫02x2 dx\int_0^2 x^2\,dx∫02​x2dx and hence show the region on the graph whose area it represents.

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Q.(a) Evaluate : $\int_0^2 x^2\,dx$ and hence show the region on the graph whose area it represents.

Unlock everything free for 14 days