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EasyIntegralsCalculus

Evaluate ∫01x2 dx\int_0^1 x^2\,dx.

The Fundamental Theorem with the power rule for antiderivatives.

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Final answer

∫01x2 dx=13≈0.33333333\int_{0}^{1} {x}^{2}\,dx = \frac{1}{3} \approx 0.33333333

How we read your problem

Evaluate the definite integral of x² from x = 0 to x = 1.

∫01x2 dx\int_{0}^{1} {x}^{2}\,dx

Step-by-step solution

  1. Start with the definite integral

    ∫01x2 dx\int_{0}^{1} {x}^{2}\,dx
  2. Power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1)

    ∫x2 dx=x33\int {x}^{2}\,dx = \frac{{x}^{3}}{3}
  3. Antiderivative

    F(x)=x33F(x) = \frac{{x}^{3}}{3}
  4. Apply the fundamental theorem of calculus

    F(1)−F(0)=0.3333333333−0=13≈0.33333333F(1) - F(0) = 0.3333333333 - 0 = \frac{1}{3} \approx 0.33333333

Formulas used

  • Power rule

    ∫xn dx=xn+1n+1+C, n≠−1\int x^n\,dx = \frac{x^{n+1}}{n+1} + C,\ n \neq -1
  • Fundamental theorem of calculus

    ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)

A definite integral gives the signed area between the curve and the x-axis: area above the axis counts as positive and area below counts as negative.

Verification

  • Differentiated the antiderivative and compared it with the integrand ✓
  • Adaptive Simpson's rule gives 0.3333333333 ✓

About integrals

Antiderivatives, substitution and the Fundamental Theorem.

Fundamental Theorem of Calculus

A definite integral equals the change in any antiderivative.

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)

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