IntegralsMedium

Definite integral — area under a curve

Calculate the area under f(x) = x² + 1 on the interval [0, 2].

Given data

f(x) = x² + 1Interval [0, 2]

Steps

0 / 4

Find an antiderivative F(x) of f(x) = x² + 1.
Write your working — the numerical result must appear in the text (logica)
Calculate F(2) — the value of the antiderivative at x = 2.
Calculate F(0) — the value of the antiderivative at x = 0.
Apply the Fundamental Theorem: ∫₀² f(x) dx = F(2) − F(0).

Full worked solution

Find an antiderivative F(x) of f(x) = x² + 1.
$F(x) = \dfrac{x^3}{3} + x$
$F(x) = \int (x^2 + 1)\,dx = \frac{x^3}{3} + x$ . The antiderivative is obtained by integrating term by term.
Calculate F(2) — the value of the antiderivative at x = 2.
$F(2) = \dfrac{2^3}{3} + 2 = \dfrac{8}{3} + 2$
$F(2) = \frac{2^3}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3} \approx \mathbf{4.667}$ .
Calculate F(0) — the value of the antiderivative at x = 0.
$F(0) = 0$
$F(0) = \frac{0^3}{3} + 0 = \mathbf{0}$ .
Apply the Fundamental Theorem: ∫₀² f(x) dx = F(2) − F(0).
$\int_0^2 (x^2 + 1)\,dx = \dfrac{14}{3} - 0 = \dfrac{14}{3}$
$\int_0^2 (x^2+1)\,dx = F(2) - F(0) = \frac{14}{3} - 0 = \mathbf{\frac{14}{3}} \approx \mathbf{4.667}$ . The area under the curve from $x=0$ to $x=2$ is $14/3$ .

Result:Area = 14/3 ≈ 4.667.

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