Area under a curve using integration

Find the finite region enclosed by y = 6 + x − x² and the x-axis. Factor:

6 + x − x² = (3 − x)(x + 2)
Roots: −2 and 3

The downward quadratic lies above the axis between those roots.

Area = ∫₋₂³(6 + x − x²) dx
= [6x + x²/2 − x³/3]₋₂³
= 27/2 − (−22/3) = 125/6

If coordinate axes have length units, the answer has square units. State any units supplied by the model.

For y = x(x − 4), the enclosed region lies below the axis from x = 0 to 4:

∫₀⁴(x² − 4x) dx
= [x³/3 − 2x²]₀⁴
= −32/3

The geometric area is 32/3. Equivalently integrate the vertical gap 0 − f(x) over this interval. A negative area is not the final geometric answer.

Let f(x) = x(x − 2)(x + 1) = x³ − x² − 2x. Its roots are −1, 0 and 2. It is positive between −1 and 0 and negative between 0 and 2.

F(x) = x⁴/4 − x³/3 − x²
∫₋₁⁰f(x) dx = 5/12
∫₀²f(x) dx = −8/3

Total area = 5/12 + 8/3 = 37/12
Signed integral = 5/12 − 8/3 = −9/4

Taking the absolute value of the single signed integral would give 9/4 and lose the cancellation that already happened. For area, make each sign-consistent piece positive before adding.

For f(x) = x(x − 2)², the finite region runs from 0 to 2. Since x ≥ 0 and the squared factor is non-negative there, the whole region is above the axis.

Area = ∫₀²(x³ − 4x² + 4x) dx
= [x⁴/4 − (4/3)x³ + 2x²]₀²
= 4/3

The curve touches the axis at the repeated root 2. Checking signs, rather than alternating them automatically at every root, prevents a false subtraction.

For f(x) = √x(6 − x), the natural domain is x ≥ 0 and the enclosed region lies above the axis from 0 to 6.

Area = ∫₀⁶(6x^1/2 − x^3/2) dx
= [4x^3/2 − (2/5)x^5/2]₀⁶
= (48/5)√6

The integrand is continuous at zero, so the endpoint is valid even though its derivative is unbounded there. The curve has maximum height 4√2 at x = 2, which helps check the sketch.

Suppose a finite region under y = f(x) has geometric area A, and track the corresponding transformed region.

A vertical factor changes every height; a horizontal factor changes every width by its reciprocal. Negative factors also reflect the region, which is why total area uses absolute values.

For f(x) = x(4 − x), the original area between 0 and 4 is 32/3. Under y = f(2x), the limits become 0 and 2 and the area is 16/3. Under y = −3f(x), the region is below the axis with area 32.

At a = 0 the graph collapses to the axis and the corresponding area is zero. The k = 0 case is not a horizontal scaling by a finite factor; the displayed reciprocal rule does not apply.

The area below y = 3x² + 2x + 1 from 0 to k is 14, with k > 0. The curve is positive, so:

∫₀^k(3x² + 2x + 1) dx = 14
k³ + k² + k − 14 = 0
(k − 2)(k² + 3k + 7) = 0

The quadratic has discriminant 9 − 28 < 0. Hence the only real solution is k = 2, which satisfies the domain. A sketch or monotonic area function can also help check uniqueness.

For f(x) = x(x − 1)(x + 4), choose a boundary a with −4 < a < 0 so that the area from a to 0 equals the area below the axis from 0 to 1.

Equal magnitudes mean the signed integral from a to 1 is zero. With F(x) = x⁴/4 + x³ − 2x²:

F(1) − F(a) = 0
a⁴ + 4a³ − 8a² + 3 = 0
(a − 1)²(a² + 6a + 3) = 0

The algebra gives a = 1 or a = −3 ± √6. The required boundary is a = −3 + √6, which lies between −4 and 0. The other negative root includes an extra negative region left of −4; a = 1 gives a zero-width integral. Neither describes the requested pair of areas.

For velocity v(t) = 12 − 4t m/s from t = 0 to 4 s, the turning time is t = 3:

∫₀³v(t) dt = 18 m
∫₃⁴v(t) dt = −2 m
Displacement = 18 − 2 = 16 m
Distance travelled = 18 + 2 = 20 m

The signed velocity integral gives displacement. Distance is the integral of speed |v|, or the sum of positive magnitudes after splitting at sign changes. A zero velocity that does not change sign need not mark a reversal.