Exercises: Area under a curve

Consider the linear function $f(x) = \frac{1}{2}x + 1$. You are asked to calculate the area of the shaded region in the graph, bounded by the function, the $OX$-axis, and the vertical lines $x = 0$ and $x = 4$. To do this:

1. Calculate the value of the area using elementary geometric formulae (area of a trapezium).

2. Calculate the definite integral $\int_{0}^{4} (\frac{1}{2}x + 1) \, dx$ using Barrow's rule and check that both results coincide.

Calculate the value of the definite integral of the polynomial function $f(x) = 3x^2 - 2x + 5$ in the interval $[1, 2]$. That is, obtain the result of:

A car wash tunnel has a metallic structure whose upper part follows the shape of the parabola $f(x) = -x^2 + 4$. Calculate the area of the flat (front) surface bounded by the parabolic structure and the ground ($OX$-axis), knowing that the support points on the ground are $x = -2$ and $x = 2$.

In physics, the distance travelled by an object can be calculated as the area under the curve in a velocity against time graph. If an e-scooter moves at a constant velocity of $v(t) = 12$ km/h during the first half hour ($t = 0.5$ h), set up and calculate the definite integral that represents the distance travelled in that time.

Find the area of the plane region bounded by the graph of the function $f(x) = x^2 - 9$ and the x-axis ($OX$).

Use the properties of the definite integral to solve the following problem:

Given that $\int_{1}^{4} f(x) \, dx = 7$ and that $\int_{4}^{6} f(x) \, dx = -3$, calculate, with justification, the value of:

1. $\int_{1}^{6} f(x) \, dx$

2. $\int_{4}^{1} 2 \times f(x) \, dx$

Calculate the total area of the region bounded by the line $f(x) = x - 2$, the $OX$-axis, and the vertical lines $x = 0$ and $x = 3$.

Note: Notice that the function crosses the $OX$-axis at a point inside the given interval, so you will need to split the region into two sub-areas.

A graphic design company wants to create a logo that includes a curved shape defined by $f(x) = \sqrt{x}$. Calculate the area of the part of the logo enclosed between this curve, the $OX$-axis, and the vertical line $x = 9$.

Calculate the area of the plane region bounded by the graph of the function $f(x) = -x^2 + 6x - 5$ and the x-axis $OX$. Sketch the region before performing the calculations to identify the x-intercepts.

You can do it on paper and send a photo with your answer.

Determine the area of the region bounded by the curve $f(x) = x^3 - 4x$ and the $OX$-axis in the interval $[-1, 2]$. Note that the function may change sign within the interval, so you will need to split the integral if necessary.

A graphic design company wants to create a logo that includes the shape of a "wave". This shape is defined by the function $f(x) = \sin(x) + 1$ in the interval $[0, \pi]$. Calculate the total area of the figure bounded by the curve, the $OX$-axis, and the vertical lines $x = 0$ and $x = \pi$.

Find the area of the region bounded by the function $f(x) = |x^2 - 1|$ and the x-axis between $x = 0$ and $x = 2$. Remember that the definition of modulus implies that the function is always non-negative, but its algebraic expression changes at the points where the argument becomes zero.

Consider the function $f(x) = x^2 - 2x + 2$. Calculate the area of the region bounded by the graph of $f(x)$, the $x$-axis, the $y$-axis, and the tangent line to the curve at the point with abscissa $x = 3$.

Find the value of a positive parameter $k$, given that the area of the region bounded by the parabola $f(x) = kx^2$ and the $OX$-axis between $x = 0$ and $x = 3$ is exactly $18$ square units.

An architect is designing the entrance of a tunnel whose front cross-section can be modeled by the function $f(x) = 4 - \frac{1}{4}x^2$, where $x$ and $y$ are expressed in meters. The floor of the tunnel coincides with the $OX$-axis. Calculate the cross-sectional area of the tunnel entrance.

Calculate the area of the region bounded by the curve $f(x) = \sqrt{x}$, the $OX$-axis, and the line $y = -x + 6$ in the first quadrant. Hint: first identify the point of intersection between the curve and the line to determine the limits of integration.

An architect is designing the side profile of a new sports hall. The main structure follows the curve of the function $f(x) = -x^2 + 6x$. To ensure stability, a metal reinforcement must be placed along the tangent line to the curve at the point with x-coordinate $x = 4$.

Calculate the area of the plane region bounded by the graph of the function $f(x)$, the mentioned tangent line, and the y-axis ($Eje\ OY$).

Consider the functions $f(x) = e^x$ and $g(x) = e^{-x}$. Let $R$ be the region bounded by the graphs of both functions and the vertical line $x = k$ (where $k > 0$).

Determine the value of the positive parameter $k$ such that the area of this region is exactly $4$ square units.

Consider the function $f(x) = |x^2 - 4|$. We want to study the region of the plane "trapped" between this curve and the horizontal line $y = 5$.

Calculate the area of the region bounded above by the line $y = 5$ and below by the graph of the function $f(x)$. Note: Take into account the points of intersection and the definition of the modulus function to split the integral if necessary.

Let the function be $f(x) = \ln(x)$. You are asked to:

1. Find the equation of the normal line to the graph of $f(x)$ at the point $P(e, 1)$.

2. Calculate the area of the plane region bounded by the curve $y = \ln(x)$, the normal line found in the previous section, and the x-axis ($Eje\ OX$).

A water tank has a cross-section whose base is defined by the $OX$-axis and its upper part by the piecewise function:

$f(x) = \begin{cases} \sqrt{x+4} & \text{if } -4 \leq x < 0 \\ 2 \cos\left(\frac{\pi x}{4}\right) & \text{if } 0 \leq x \leq 2 \end{cases}$

Calculate the total cross-sectional area of the tank by computing the definite integral in each interval.

Determine the value of the constant $a \in \mathbb{R}$ with $a > 1$, knowing that the area of the region bounded by the curve $y = \frac{1}{x^2}$, the $x$-axis, and the vertical lines $x = 1$ and $x = a$ is equal to twice the area of the region bounded by the same curve $y = \frac{1}{x^2}$, the $x$-axis, and the lines $x = a$ and $x = a^2$.

Find the area of the region bounded by the function $f(x) = x^3 - 3x^2 + 2x$ and the horizontal line passing through the local maximum of this function.

Hint: You must first locate the local maximum using the derivative, find the y-coordinate at that point, and then determine the points of intersection between the curve and the resulting line.

Let the region $A$ be the region in the first quadrant bounded by the parabola $y = x^2$ and the line $y = 9$. There is a vertical line with equation $x = c$ (with $0 < c < 3$) that divides the region $A$ into two subregions of identical area.

Calculate the exact value of the constant $c$.