Math / Calculus

QuestionUse a definite integral to find the area under the graph of $y=-2 x^{2}+3 x+4$ between $x=0$ and $x=2$ . Round to 2 decimal places. eTextbook and Media

Studdy Solution

STEP 1

What is this asking? We need to find the area under a parabola between two vertical lines, using integrals! Watch out! Remember the power rule for integration and don't forget to plug in the upper and lower limits of integration correctly.
Also, watch those signs!

STEP 2

1. Define the function
2. Find the indefinite integral
3. Calculate the definite integral

STEP 3

Our function is $y = -2x^2 + 3x + 4$ , and we're looking at it between $x = 0$ and $x = 2$ .
Let's call this function $f(x)$ , so $f(x) = -2x^2 + 3x + 4$ .

STEP 4

To find the area under the curve, we need to find the definite integral of $f(x)$ from $x = 0$ to $x = 2$ .
But first, let's find the indefinite integral!

STEP 5

Remember the power rule for integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ , where $C$ is the constant of integration.

STEP 6

Applying the power rule to each term in $f(x)$ , we get: $\int (-2x^2 + 3x + 4) \, dx = -2 \cdot \frac{x^3}{3} + 3 \cdot \frac{x^2}{2} + 4x + C$ $\int (-2x^2 + 3x + 4) \, dx = -\frac{2}{3}x^3 + \frac{3}{2}x^2 + 4x + C$ So, our indefinite integral is $-\frac{2}{3}x^3 + \frac{3}{2}x^2 + 4x + C$ .

STEP 7

Now, let's evaluate the definite integral from $x = \textbf{0}$ to $x = \textbf{2}$ .
We denote this as: $\int_{0}^{2} (-2x^2 + 3x + 4) \, dx$

STEP 8

We plug in the upper limit $x = 2$ into our indefinite integral: $-\frac{2}{3}(\textbf{2})^3 + \frac{3}{2}(\textbf{2})^2 + 4(\textbf{2}) + C = -\frac{16}{3} + 6 + 8 + C = -\frac{16}{3} + 14 + C$

STEP 9

Next, we plug in the lower limit $x = 0$ : $-\frac{2}{3}(\textbf{0})^3 + \frac{3}{2}(\textbf{0})^2 + 4(\textbf{0}) + C = 0 + C$

STEP 10

Finally, we subtract the result from plugging in the lower limit from the result of plugging in the upper limit: $\left( -\frac{16}{3} + 14 + C \right) - (C) = -\frac{16}{3} + 14 = \frac{-16 + 42}{3} = \frac{26}{3}$ Notice how the constant of integration, $C$ , adds to zero when we subtract!

STEP 11

As a decimal rounded to two places, this is approximately $\frac{26}{3} \approx 8.67$ .

STEP 12

The area under the graph of $y = -2x^2 + 3x + 4$ between $x = 0$ and $x = 2$ is approximately 8.67.

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