Evaluate the integral $\int_0^1(2x^2-x^3)dx$

The correct answer is Option d: $\frac{5}{12}$. We are asked to evaluate the definite integral: $$\int_0^1 (2x^2 - x^3) dx$$ Using the power rule of integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$, we find the antiderivative: $$\int (2x^2 - x^3) dx = 2 \left(\frac{x^3}{3}\right) - \frac{x^4}{4} = \frac{2x^3}{3} - \frac{x^4}{4}$$ Now we apply the Fundamental Theorem of Calculus by substituting the upper limit $1$ and lower limit $0$: $$\int_0^1 (2x^2 - x^3) dx = \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^1$$ $$= \left( \frac{2(1)^3}{3} - \frac{1^4}{4} \right) - \left( \frac{2(0)^3}{3} - \frac{0^4}{4} \right)$$ $$= \left( \frac{2}{3} - \frac{1}{4} \right) - 0$$ Find a common denominator to subtract the fractions: $$\frac{2}{3} - \frac{1}{4} = \frac{2 \times 4 - 1 \times 3}{12} = \frac{8 - 3}{12} = \frac{5}{12}$$ Hence, **Option D** is the correct answer.

Basic Applications of CalculusPYQ Jan 26Question 4271 of 28

All Questions

Evaluate the integral $\displaystyle \int_0^1(2x^2-x^3)dx$

Options

\displaystyle \frac{1}{3}

\displaystyle \frac{4}{3}

\displaystyle \frac{7}{12}

\displaystyle \frac{5}{12}

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Correct Answer

✅ Option d — $\displaystyle \frac{5}{12}$

All Options:

A $\displaystyle \frac{1}{3}$
B $\displaystyle \frac{4}{3}$
C $\displaystyle \frac{7}{12}$
D $\displaystyle \frac{5}{12}$

Detailed Solution & Explanation

We are asked to evaluate the definite integral:

\int_0^1 (2x^2 - x^3) dx

Using the power rule of integration,

\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1}

, we find the antiderivative:

\int (2x^2 - x^3) dx = 2 \left(\frac{x^3}{3}\right) - \frac{x^4}{4} = \frac{2x^3}{3} - \frac{x^4}{4}

Now we apply the Fundamental Theorem of Calculus by substituting the upper limit

\displaystyle 1

and lower limit

\displaystyle 0

\int_0^1 (2x^2 - x^3) dx = \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^1

= \left( \frac{2(1)^3}{3} - \frac{1^4}{4} \right) - \left( \frac{2(0)^3}{3} - \frac{0^4}{4} \right)

= \left( \frac{2}{3} - \frac{1}{4} \right) - 0

Find a common denominator to subtract the fractions:

\frac{2}{3} - \frac{1}{4} = \frac{2 \times 4 - 1 \times 3}{12} = \frac{8 - 3}{12} = \frac{5}{12}

Hence, **Option D** is the correct answer.

About This Chapter: Basic Applications of Calculus

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Limits, Continuity, Derivatives, Integrals

This chapter covers Limits, Continuity, Derivatives, Integrals and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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Evaluate the integral ∫01(2x2−x3)dx\displaystyle \int_0^1(2x^2-x^3)dx∫01​(2x2−x3)dx