# Area Under a Curve – Examples with Answers

The area under a curve can be found using definite integrals. In turn, the definite integrals are calculated by integrating the function and evaluating both the lower limit and the upper limit. The lower limit is subtracted from the upper limit to obtain a given value for the area.

Here, we will learn how to find the area under a curve. We will look at a few solved exercises of the area under a curve. In addition, we will explore some practice problems to apply what has been learned.

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Learning to find the area under a curve with examples.

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##### CALCULUS

Relevant for

Learning to find the area under a curve with examples.

See examples

## Process used to find the area under a curve

We consider the area $latex A$ under the curve $latex f(x)$ shown in the following diagram:

We can find the area under this curve using a definite integral. In this case, the area under the curve is represented by $latex A= \int_{a}^{b} f(x)dx$, where,

• $latex dx$ indicates that the limits $latex a$ and $latex b$ are limits of x.
• The constant $latex a$ is the lower limit of the integral.
• The constant $latex b$ is the upper limit of the integral.

Taking this into account, we can follow the steps below to find the area under a curve, assuming we want to find the area under $latex 2x$ between $latex x=0$ and $latex x=1$.

Step 1: Form a definite integral with the given information. In this case, we have the following integral:

$latex \int_{0}^{1} 2xdx$

Step 2: Obtain the integral of the function and express it using brackets, where we write the limits of integration as follows:

$latex \int_{0}^{1} 2xdx=[x^2+c]_{0}^{1}$

Step 3: Evaluate the upper and lower limits on the integrated expression. We subtract the lower limit from the upper limit:

$latex [x^2+c]_{0}^{1}=[(1)^2+c]-[(0)^2+c]$

Step 4: Simplify to a single numeric value:

$latex =[(1)^2+c]-[(0)^2+c]$

$latex =[1+c]-[0+c]$

$latex =1$

The value found corresponds to the area.

When solving definite integrals, the constants of integration are generally ignored, since they will be canceled in step 3 anyway.

## Worked examples of the area under a curve

### EXAMPLE 1

What is the area under the curve represented by $latex y=x^2$ between $latex x=1$ and $latex x=3$?

To solve this problem, we have to start by forming a definite integral with the given information. Therefore, we have:

$$A=\int_{1}^{3} x^2 dx$$

Now, we find the integral of the expression and keep the limits of integration using square brackets:

$$A=\left[ \frac{x^3}{3} \right]_{1}^{3}$$

When we evaluate the limits of integration, we have:

$$A=\left[ \frac{(3)^3}{3} \right]-\left[ \frac{(1)^3}{3} \right]$$

Finally, we can simplify:

$$A=\left[ \frac{27}{3} \right]-\left[ \frac{1}{3} \right]$$

$$A=-\left[ \frac{1}{3} \right]$$

$$A= \frac{26}{3}$$

### EXAMPLE 2

Determine the area under the curve that is represented by $latex y=\frac{1}{3}x^2+2$ since $latex x=0$ and $latex x=3$.

With the given information, we can form the following definite integral:

$$A=\int_{0}^{3} \frac{1}{3} x^2+2 dx$$

Now, we integrate the given expression and keep the limits of integration:

$$A=\left[ \frac{1}{9} x^3+2x \right]_{0}^{3}$$

When we evaluate the limits of integration in the integrated expression, we have:

$$A=\left[\frac{1}{9}(3)^3+2(3) \right]-\left[\frac{1}{9}(0)^3+2(0) \right]$$

Simplifying, we have:

$$A=[3+6]-$$

$$A=9$$

### EXAMPLE 3

What is the area of the curve represented by $latex y=x^3-4x$ from $latex x=-2$ to $latex x=0$?

Start by forming a definite integral with the given information:

$$A=\int_{-2}^{0} x^3-4x dx$$

Now, we integrate the expression and keep the limits of integration:

$$A=\left[ \frac{x^4}{4} -2x^2 \right]_{-2}^{0}$$

When evaluating the limits of integration, we have:

$$A=\left[ \frac{(0)^4}{4} -2(0)^2 \right]-\left[ \frac{(-2)^4}{4} -2(-2)^2 \right]$$

Simplify to obtain a single value for the area:

$$A=[ 0 -0]-\left[ \frac{16}{4} -2(4) \right]$$

$latex A=-[ 4-8]$

$latex A= 4$

### EXAMPLE 4

Find the area under the curve $latex y=x^2+x+2$ from $latex x=-1$ to $latex x=2$.

Forming an integral with the given information, we have:

$$A=\int_{-1}^{2} x^2+x+2 dx$$

Now, we can integrate the expression while keeping the limits of integration:

$$A=\left[ \frac{x^3}{3} +\frac{x^2}{2}+2x \right]_{-1}^{2}$$

Evaluating this, we have:

$$A=\left[ \frac{(2)^3}{3} +\frac{(2)^2}{2}+2(2) \right]-\left[ \frac{(-1)^3}{3} +\frac{(-1)^2}{2}+2(-1) \right]$$

When we simplify, we have:

$$A=\left[ \frac{8}{3} +2+4 \right]-\left[ -\frac{1}{3} +\frac{1}{2}-2 \right]$$

$$A=\left[ \frac{26}{3}\right]-\left[ -\frac{11}{6} \right]$$

$latex A=\frac{63}{6}$

$latex A= 10~\frac{1}{2}$

### EXAMPLE 5

If we have a curve represented by $latex y=x^2+4x$, find the area of the region from $latex x=-2$ to $latex x=0$.

In this case, the region we want is below the x-axis. However, we can follow the same process and find a definite integral with the given information:

$$A=\int_{-2}^{0} x^2 +4x dx$$

Determining the integral of the expression and maintaining the limits of integration, we have:

$$A=\left[ \frac{x^3}{3} +2x^2 \right]_{-2}^{0}$$

Evaluating the limits of integration, we have:

$$A=\left[ \frac{(0)^3}{3} +2(0)^2 \right]-\left[ \frac{(-2)^3}{3} +2(-2)^2 \right]$$

When simplifying, we get the following:

$$A=[ 0 +0]-\left[ \frac{-8}{3} +8 \right]$$

$$A=-\left[ \frac{16}{3} \right]$$

We see that we got a negative value. The reason for this is that the area we want is under the x-axis. So, we ignore the minus sign and the area is equal to $latex A= \frac{16}{3}$.

In this example, we can understand why it is useful to draw a simple graph when we want to find the area under a curve.

### EXAMPLE 6

Find the area under the curve $latex y=3x^2-3x-6$ from $latex x=-1$ to $latex x=2$.

This example is similar to the previous one. Therefore, we can use the same process and ignore the resulting negative sign.

Forming the definite integral, we have:

$$A=\int_{-1}^{2} 3x^2-3x-6 dx$$

Determining the integral of the expression and maintaining the limits of integration, we have:

$$A=\left[ x^3 -\frac{3x^2}{2}-6x \right]_{-1}^{2}$$

When we evaluate the limits of integration, we have:

$$A=\left[ (2)^3 -\frac{3(2)^2}{2}-6(2) \right]-\left[ (-1)^3 -\frac{3(-1)^2}{2}-6(-1) \right]$$

Simplifying, we have:

$$A=[ 8 -6-12 ]-\left[ -1 -\frac{3}{2}+6 \right]$$

$$A=[ -10 ]-\left[ \frac{7}{2} \right]$$

$$A=-\frac{27}{2}=-13.5$$

The area is equal to $latex A= 13.5$.

### EXAMPLE 7

Find the area under the curve $latex y=2x^3-8x$ from $latex x=-2$ to $latex x=2$.

We can solve this by finding the areas below the x-axis and above the x-axis separately. However, in this case, we see that the graph is symmetric, and both regions have the same area.

Therefore, we can find the area of one region and then multiply it by 2. Thus, we start by forming the following definite integral:

$$A=\int_{-2}^{0} 2x^3-8x dx$$

Now, we write as follows:

$$A=\left[ \frac{x^4}{2}-4x^2 \right]_{-2}^{0}$$

Evaluating this, we have:

$$A=\left[ \frac{(0)^4}{2}-4(0)^2 \right]-\left[ \frac{(-2)^4}{2}-4(-2)^2 \right]$$

Finally, we simplify to obtain a single value:

$latex A=[0-0]-[ 8 -16]$

$latex A= 8$

This means that the area of the region from $latex x=-2$ to $latex x=2$ is 16.

### EXAMPLE 8

If we have the curve $latex y=x^3-4x^2+3x$, what is the area under the curve from $latex x=0$ to $latex x=3$?

In this case, the required area has two parts, $latex A_{1}$, the area above the x-axis, and $latex A_{2}$, the area below the x-axis. So, let’s find these areas separately.

For the area $latex A_{1}$, we have the following definite integral:

$$A_{1}=\int_{0}^{1} (x^3-4x^2+3x) dx$$

We can solve this integral as follows:

$$A_{1}=\left[ \frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2} \right]_{0}^{1}$$

$$A_{1}=\left[ \frac{1}{4}-\frac{4}{3}+\frac{3}{2} \right]-[ 0]$$

$$A_{1}=\frac{5}{12}$$

For the area $latex A_{2}$, we have the following definite integral:

$$A_{2}=\int_{1}^{3} (x^3-4x^2+3x) dx$$

And we solve as follows:

$$A_{2}=\left[ \frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2} \right]_{1}^{3}$$

$$A_{2}=\left[ \frac{81}{4}-36+\frac{27}{2} \right]-\left[ \frac{1}{4}-\frac{4}{3}+\frac{3}{2} \right]$$

$$A_{2}=-\frac{9}{4}-\frac{5}{12}$$

$$A_{2}=-\frac{8}{3}$$

Finally, we calculate the total area $latex A$ as follows:

$$A=A_{1}+A_{2}$$

$$A=\frac{5}{12}+\frac{8}{3}$$

$$A=\frac{37}{12}$$

## Area under a curve – Practice problems

Area under a curve quiz  You have completed the quiz!

#### Find the area under the curve $latex y=2x^2+3$ from $latex x=-1$ to $latex x=2$.

Write the answer in the input box.

$latex A=$

Interested in learning more about integrals of functions? You can look at these pages: ### Jefferson Huera Guzman

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  