Integration: Learning Objectives 1, 2, 3, 4

The width of each rectangle is $\Delta x = \frac{b-a}{n}$. For part (a), the height of each rectangle is determined by the function value at its right endpoint. For part (b), the height is determined by the left endpoint. Consider the shape of the function $f(x)=1/x$ on the interval to determine if the estimates are over or under.

1. $R_4 \approx 0.6345$. This is an underestimate because $f(x)=1/x$ is a decreasing function on the interval $[1,2]$.
2. $L_4 \approx 0.7595$. This is an overestimate.

The width of each rectangle is $\Delta x = \frac{2-1}{4} = 0.25$.

a) Right Endpoints ($R_4$): The endpoints are $1.25, 1.5, 1.75, 2$.
The area is: \[ R_4 = \Delta x [f(1.25) + f(1.5) + f(1.75) + f(2)] \] \[ = 0.25 \left( \frac{1}{1.25} + \frac{1}{1.5} + \frac{1}{1.75} + \frac{1}{2} \right) \] \[ = 0.25 \left( 0.8 + 0.666... + 0.5714... + 0.5 \right) \approx 0.25(2.538) \approx 0.6345 \] Since $f(x) = 1/x$ is decreasing, the right endpoint is the minimum value in each subinterval, making this an underestimate.

b) Left Endpoints ($L_4$): The endpoints are $1, 1.25, 1.5, 1.75$.
The area is: \[ L_4 = \Delta x [f(1) + f(1.25) + f(1.5) + f(1.75)] \] \[ = 0.25 \left( \frac{1}{1} + \frac{1}{1.25} + \frac{1}{1.5} + \frac{1}{1.75} \right) \] \[ = 0.25 \left( 1 + 0.8 + 0.666... + 0.5714... \right) \approx 0.25(3.038) \approx 0.7595 \] Since $f(x) = 1/x$ is decreasing, the left endpoint is the maximum value in each subinterval, making this an overestimate.

Use the integral property $\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx$.

5

We can split the integral from 0 to 10 at the point 8: \[ \int_0^{10} f(x) \,dx = \int_0^{8} f(x) \,dx + \int_8^{10} f(x) \,dx \] Substitute the known values: \[ 17 = 12 + \int_8^{10} f(x) \,dx \] Solve for the unknown integral: \[ \int_8^{10} f(x) \,dx = 17 - 12 = 5 \]

Graph the function $y = 1-x$. The definite integral represents the net area between the graph and the x-axis from $x=-1$ to $x=2$. This area consists of a triangle above the axis and a triangle below the axis.

$\frac{3}{2}$

The graph of $y=1-x$ is a straight line that crosses the x-axis at $x=1$.

1. Area of Triangle 1 (above x-axis): From $x=-1$ to $x=1$, we have a triangle with base $1 - (-1) = 2$ and height $f(-1) = 1 - (-1) = 2$. Its area is $A_1 = \frac{1}{2}(2)(2) = 2$.

2. Area of Triangle 2 (below x-axis): From $x=1$ to $x=2$, we have a triangle with base $2 - 1 = 1$ and height $|f(2)| = |1 - 2| = 1$. Its area is $A_2 = \frac{1}{2}(1)(1) = 0.5$.

3. Net Area: The definite integral is the area above the axis minus the area below the axis. \[ \int_{-1}^2 (1-x)\,dx = A_1 - A_2 = 2 - 0.5 = 1.5 = \frac{3}{2} \]

The area under a non-negative function $f(x)$ from $a$ to $b$ is given by the definite integral $\int_a^b f(x) \, dx$. Use the Fundamental Theorem of Calculus.

$\frac{1}{3}$

The area is given by the definite integral: \[ \text{Area} = \int_0^1 x^2 \, dx \] Using the Fundamental Theorem of Calculus, we find the antiderivative of $x^2$, which is $\frac{x^3}{3}$. \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3} \]

Consider the conditions required to use the Fundamental Theorem of Calculus. Is the function $f(x) = \frac{1}{x^2}$ continuous on the interval of integration $[-1, 3]$?

The calculation is wrong because the Fundamental Theorem of Calculus was applied to a function that is not continuous on the interval of integration. The integral does not exist (it diverges).

The Fundamental Theorem of Calculus, which allows us to evaluate $\int_a^b f(x) dx$ using an antiderivative, requires that the function $f(x)$ be continuous on the closed interval $[a, b]$.

In this case, the function is $f(x) = \frac{1}{x^2}$. This function has an infinite discontinuity at $x=0$.

The interval of integration is $[-1, 3]$, which contains the point of discontinuity $x=0$.

Since the function is not continuous on the entire interval, the Fundamental Theorem of Calculus cannot be applied as shown. The integral is an improper integral, and in this case, it diverges to infinity. The negative result obtained is meaningless.

Find the antiderivative of the polynomial term-by-term and then evaluate it at the upper and lower bounds using the Fundamental Theorem of Calculus.

$\frac{26}{3}$

1. Find the antiderivative: \[ \int (x^2 + 2x - 4) \, dx = \frac{x^3}{3} + \frac{2x^2}{2} - 4x = \frac{x^3}{3} + x^2 - 4x \] 2. Apply the Fundamental Theorem of Calculus: \[ \left[ \frac{x^3}{3} + x^2 - 4x \right]_1^3 = \left(\frac{3^3}{3} + 3^2 - 4(3)\right) - \left(\frac{1^3}{3} + 1^2 - 4(1)\right) \] \[ = \left(\frac{27}{3} + 9 - 12\right) - \left(\frac{1}{3} + 1 - 4\right) \] \[ = (9 + 9 - 12) - \left(\frac{1}{3} - 3\right) = 6 - \left(-\frac{8}{3}\right) = 6 + \frac{8}{3} = \frac{18}{3} + \frac{8}{3} = \frac{26}{3} \]

First, simplify the integrand by dividing each term in the numerator by $\sqrt{x} = x^{1/2}$. Then use the power rule for integration.

$\frac{82}{5}$

1. Simplify the integrand: \[ \frac{2+x^2}{x^{1/2}} = \frac{2}{x^{1/2}} + \frac{x^2}{x^{1/2}} = 2x^{-1/2} + x^{3/2} \] 2. Find the antiderivative: \[ \int (2x^{-1/2} + x^{3/2}) \, dx = 2\frac{x^{1/2}}{1/2} + \frac{x^{5/2}}{5/2} = 4x^{1/2} + \frac{2}{5}x^{5/2} \] 3. Apply the Fundamental Theorem of Calculus: \[ \left[ 4\sqrt{x} + \frac{2}{5}x^{5/2} \right]_1^4 = \left(4\sqrt{4} + \frac{2}{5}(4)^{5/2}\right) - \left(4\sqrt{1} + \frac{2}{5}(1)^{5/2}\right) \] \[ = \left(4(2) + \frac{2}{5}(32)\right) - \left(4 + \frac{2}{5}\right) = \left(8 + \frac{64}{5}\right) - \left(\frac{20}{5} + \frac{2}{5}\right) \] \[ = \left(\frac{40+64}{5}\right) - \frac{22}{5} = \frac{104}{5} - \frac{22}{5} = \frac{82}{5} \]

Rewrite the function as $f(x) = \frac{1}{x^4}$ and check for any points of discontinuity within the interval $[-2, 1]$.

The integral does not exist (it diverges) because the function $f(x) = x^{-4}$ is not continuous on the interval $[-2, 1]$. The Fundamental Theorem of Calculus cannot be applied.

The function $f(x) = x^{-4} = \frac{1}{x^4}$ has an infinite discontinuity at $x=0$.

The interval of integration is $[-2, 1]$, which contains the point of discontinuity $x=0$.

Because the function is not continuous over the entire interval, the Fundamental Theorem of Calculus cannot be applied. The integral is an improper integral that diverges, so it does not have a finite value.

Recall the integration rule for exponential functions with a base other than $e$: $\int a^x \, dx = \frac{a^x}{\ln(a)} + C$.

$\frac{9}{\ln(10)}$

The antiderivative of $10^x$ is $\frac{10^x}{\ln(10)}$.

Using the Fundamental Theorem of Calculus: \[ \int_{0}^1 10^x \, dx = \left[ \frac{10^x}{\ln(10)} \right]_0^1 \] \[ = \frac{10^1}{\ln(10)} - \frac{10^0}{\ln(10)} = \frac{10}{\ln(10)} - \frac{1}{\ln(10)} = \frac{9}{\ln(10)} \]

This is an indefinite integral. Integrate each term separately using the power rule and the rule for integrating the natural exponential function. Don't forget the constant of integration, $C$.

$x^2 - e^x + C$

We integrate term-by-term: \[ \int (2x - e^x) \, dx = \int 2x \, dx - \int e^x \, dx \] Apply the power rule to the first term and the exponential rule to the second: \[ = 2\left(\frac{x^2}{2}\right) - e^x + C \] \[ = x^2 - e^x + C \]