Integration — practice questions

We integrate the function term by term.

$\int \left( 4x^3 + \frac{6}{x} - 3e^{2x} \right) \, dx$

1. Integrate $4x^3$: Using the power rule, $\int 4x^3 \, dx = 4 \times \frac{x^{3+1}}{3+1} = 4 \frac{x^4}{4} = x^4$.
2. Integrate $\frac{6}{x}$: This is $6 \times \frac{1}{x}$. The integral is $6 \ln|x|$.
3. Integrate $-3e^{2x}$: Using the rule for exponentials, $\int -3e^{2x} \, dx = -3 \times \frac{1}{2}e^{2x} = -\frac{3}{2}e^{2x}$.
4. Combine and add the constant of integration: Don't forget to add $+c$ at the end.

So, $\int \left( 4x^3 + \frac{6}{x} - 3e^{2x} \right) \, dx = x^4 + 6\ln|x| - \frac{3}{2}e^{2x} + c$.

The region is bounded by $x=0$ and $x=\frac{\pi}{2}$. The function $y=\sin(2x)$ is non-negative on this interval.

We need to evaluate the definite integral: Area $= \int_0^{\pi/2} \sin(2x) \, dx$

1. Find the indefinite integral: $\int \sin(2x) \, dx = -\frac{1}{2}\cos(2x)$.
2. Apply the limits: We use square brackets to denote the evaluation. Area $= \left[ -\frac{1}{2}\cos(2x) \right]_0^{\pi/2}$
3. Substitute the upper limit ($b = \pi/2$): $-\frac{1}{2}\cos(2 \times \frac{\pi}{2}) = -\frac{1}{2}\cos(\pi) = -\frac{1}{2}(-1) = \frac{1}{2}$.
4. Substitute the lower limit ($a = 0$): $-\frac{1}{2}\cos(2 \times 0) = -\frac{1}{2}\cos(0) = -\frac{1}{2}(1) = -\frac{1}{2}$.
5. Calculate $F(b) - F(a)$: Area $= (\frac{1}{2}) - (-\frac{1}{2}) = \frac{1}{2} + \frac{1}{2} = 1$.

The exact area is 1 square unit.

1. Identify parameters: The interval is $[a, b] = [1, 5]$. The number of strips is $n=4$.
2. Calculate strip width, $h$: $h = \frac{b-a}{n} = \frac{5-1}{4} = 1$.
3. Determine x-values: The x-values (ordinates) will be $x_0=1, x_1=2, x_2=3, x_3=4, x_4=5$.
4. Calculate corresponding y-values: Let $y(x) = \frac{x}{x+1}$. $y_0 = y(1) = \frac{1}{1+1} = 0.5$ $y_1 = y(2) = \frac{2}{2+1} = \frac{2}{3} \approx 0.66667$ $y_2 = y(3) = \frac{3}{3+1} = 0.75$ $y_3 = y(4) = \frac{4}{4+1} = 0.8$ $y_4 = y(5) = \frac{5}{5+1} = \frac{5}{6} \approx 0.83333$
5. Apply the Trapezium Rule formula: $\int_1^5 y \, dx \approx \frac{h}{2}[y_0 + y_4 + 2(y_1 + y_2 + y_3)]$ $\approx \frac{1}{2}[0.5 + 0.83333 + 2(0.66667 + 0.75 + 0.8)]$ $\approx 0.5[1.33333 + 2(2.21667)]$ $\approx 0.5[1.33333 + 4.43334]$ $\approx 0.5[5.76667] \approx 2.883335$

Rounding to 3 decimal places, the approximation is 2.883.