Question

find the area under the curve \\(f(x) = x^2 + 2x\\) between \\(x = 1\\) and \\(x = 4\\).

Explanation:

Set up the integral

We find the area using a definite integral.

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

Here, \(f(x) = x^2 + 2x\), \(a = 1\), and \(b = 4\).

$$ A = \int_{1}^{4} (x^2 + 2x) \, dx $$

Find the antiderivative

We apply the power rule for integration.

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

The antiderivative \(F(x)\) is:

$$ F(x) = \frac{x^3}{3} + x^2 $$

Evaluate at the upper limit

We substitute \(x = 4\) into \(F(x)\).

$$ F(4) = \frac{4^3}{3} + 4^2 = \frac{64}{3} + 16 $$

Convert to a common denominator:

$$ F(4) = \frac{64}{3} + \frac{48}{3} = \frac{112}{3} $$

Evaluate at the lower limit

We substitute \(x = 1\) into \(F(x)\).

$$ F(1) = \frac{1^3}{3} + 1^2 = \frac{1}{3} + 1 $$

Convert to a common denominator:

$$ F(1) = \frac{1}{3} + \frac{3}{3} = \frac{4}{3} $$

Calculate the final area

We subtract \(F(1)\) from \(F(4)\).

$$ A = F(4) - F(1) = \frac{112}{3} - \frac{4}{3} $$

Simplify the fraction:

$$ A = \frac{108}{3} = 36 $$

Answer:

36