Find the area under the curve y = x2 from x = 0 to x = 3. Find the area under y = 2x + 1 from x = 1 to x = 4…

Question

Find the area under the curve y = x2 from x = 0 to x = 3. Find the area under y = 2x + 1 from x = 1 to x = 4. Find the area between y = x2 and y = x from x = 0 to x = 1.

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Right 🌹CONFIDENTIALITY, let's tackle these area problems.

Q1. Find the area under the curve $y = x^2$ from $x = 0$ to $x = 3$ .

Set up the definite integral for the area.
$A = \int_0^3 x^2 \, dx$
Find the antiderivative of $x^2$ .
$\int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$
Evaluate the antiderivative at the upper and lower limits using the Fundamental Theorem of Calculus.
$A = \left[\frac{x^3}{3}\right]_0^3 = \left(\frac{3^3}{3}\right) - \left(\frac{0^3}{3}\right)$
Calculate the numerical value.
$A = \frac{27}{3} - 0 = 9$
The final answer is $\boxed{9}$ .

Q2. Find the area under $y = 2x + 1$ from $x = 1$ to $x = 4$ .
Set up the definite integral for the area.
$A = \int_1^4 (2x + 1) \, dx$
Find the antiderivative of $2x + 1$ .
$\int (2x + 1) \, dx = 2 \cdot \frac{x^2}{2} + x = x^2 + x$
Evaluate the antiderivative at the upper and lower limits.
$A = [x^2 + x]_1^4 = ((4)^2 + 4) - ((1)^2 + 1)$
Calculate the numerical value.
$A = (16 + 4) - (1 + 1) = 20 - 2 = 18$
The final answer is $\boxed{18}$ .

Q3. Find the area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$ .
Determine which function is greater over the interval $[0, 1]$.
For $x \in (0, 1)$ , $x > x^2$ . For example, if $x=0.5$ , then $x=0.5$ and $x^2=0.25$ . So $y=x$ is the upper curve and $y=x^2$ is the lower curve.
Set up the definite integral for the area between the curves.
$A = \int_0^1 (x - x^2) \, dx$
Find the antiderivative of $x - x^2$ .
$\int (x - x^2) \, dx = \frac{x^2}{2} - \frac{x^3}{3}$
Evaluate the antiderivative at the upper and lower limits.
$A = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \left(\frac{1^2}{2} - \frac{1^3}{3}\right) - \left(\frac{0^2}{2} - \frac{0^3}{3}\right)$
Calculate the numerical value.
$A = \left(\frac{1}{2} - \frac{1}{3}\right) - (0 - 0) = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$
The final answer is $\boxed{\frac{1}{6}}$ .

Find the area under the curve y = x2 from x = 0 to x = 3. Find the area under y = 2x + 1 from x = 1 to x = 4. Find the area between y = x2 and y = x from x = 0 to x = 1.

Step-by-step solution

Set up the definite integral for the area.

Find the antiderivative of $x^2$ .

Evaluate the antiderivative at the upper and lower limits using the Fundamental Theorem of Calculus.

Calculate the numerical value.

Set up the definite integral for the area.

Find the antiderivative of $2x + 1$ .

Evaluate the antiderivative at the upper and lower limits.

Calculate the numerical value.

Determine which function is greater over the interval $[0, 1]$.

Set up the definite integral for the area between the curves.

Find the antiderivative of $x - x^2$ .

Evaluate the antiderivative at the upper and lower limits.

Calculate the numerical value.

Still stuck on this one?

More Mathematics Questions

Find the area under the curve y = x2 from x = 0 to x = 3. Find the area under y = 2x + 1 from x = 1 to x = 4. Find the area between y = x2 and y = x from x = 0 to x = 1.

Step-by-step solution

Set up the definite integral for the area.

Find the antiderivative of x2x^2x2.

Evaluate the antiderivative at the upper and lower limits using the Fundamental Theorem of Calculus.

Calculate the numerical value.

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Find the antiderivative of 2x+12x + 12x+1.

Evaluate the antiderivative at the upper and lower limits.

Calculate the numerical value.

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Find the antiderivative of x−x2x - x^2x−x2.

Evaluate the antiderivative at the upper and lower limits.

Calculate the numerical value.

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