If I_1=_0^12^x^2 dx, I_2=_0^12^x^3 dx, I_3=_1^22^x^2 dx and I_4=_1^22^x^3 dx, then

D. I_4>I_3. Step 1: On (0,1), x^3 x^2, so 2^x^3>2^x^2 and I_4>I_3. Check: At x=1/2: 1/8 2.25. Numerically I_3 6.05 and I_4 35.9. Trap: (B) I_2>I_1 and (C) I_3>I_4 come from assuming x^3>x^2 everywhere; that holds only for x>1.

Question 9 of 31 · Definite Integration

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NIMCET 2012

If $I_{1} = \int_{0}^{1} 2^{x^{2}} d x$ , $I_{2} = \int_{0}^{1} 2^{x^{3}} d x$ , $I_{3} = \int_{1}^{2} 2^{x^{2}} d x$ and $I_{4} = \int_{1}^{2} 2^{x^{3}} d x$ , then

Answer (Detailed Solution Below)View answer ▾

Answer: Option D — $I_{4} > I_{3}$

Detailed Solution

On $(0, 1)$ , $x^{3} < x^{2}$ , so $2^{x^{3}} < 2^{x^{2}}$ and $I_{2} < I_{1}$ .

On $(1, 2)$ , $x^{3} > x^{2}$ , so $2^{x^{3}} > 2^{x^{2}}$ and $I_{4} > I_{3}$ .

✓ Check

x = \frac{1}{2}

\frac{1}{8} < \frac{1}{4}

; at

x = \frac{3}{2}

3.375 > 2.25

. Numerically

I_{3} \approx 6.05

and

I_{4} \approx 35.9

⚠ Common trap

(B)

I_{2} > I_{1}

and (C)

I_{3} > I_{4}

come from assuming

x^{3} > x^{2}

everywhere; that holds only for

x > 1

Quick tip

$x^{3} < x^{2}$ on $(0, 1)$ and $x^{3} > x^{2}$ on $(1, 2)$

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Definite Integration — Preparation

This question tests your understanding of Definite Integration, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A definite integral gives a number rather than a family of functions. It is evaluated using the fundamental theorem of calculus: the integral of f(x) from a to b equals F(b) - F(a), where F is an antiderivative of f. This topic covers evaluating definite integrals, substitution with a change of limits, integration by parts with limits, and the definite integral as the limit of a sum. The properties of definite integrals are the key to most exam questions. The most useful is the "a + b - x" (or "a - x") property, which turns many difficult integrals into simple ones by adding the integral to itself. Also remember that the integral of an odd function over a symmetric interval from -a to a is zero, while the integral of an even function equals twice the integral from 0 to a. Practise the definite integration questions below, each with a step-by-step solution showing which property is used.

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