Comprehensive Exercises in the Integral Calculus of One Variable

Approach:

Use the linearity of integration to split the polynomial into term-by-term integrals.

Steps:

1. $\int (3x^2 - 2x + 5) dx = 3\int x^2 dx - 2\int x dx + 5\int dx$
2. $= 3 \cdot \frac{x^3}{3} - 2 \cdot \frac{x^2}{2} + 5x + C$
3. $= x^3 - x^2 + 5x + C$

Answer: $x^3 - x^2 + 5x + C$

Approach:

Notice that the numerator $2x$ is exactly the derivative of the denominator $1 + x^2$, so the differential can be worked.

Steps:

1. Put $u = 1 + x^2$, so $du = 2x dx$
2. $\int \frac{2x}{1 + x^2} dx = \int \frac{1}{u} du$
3. $= \ln|u| + C = \ln(1 + x^2) + C$

Answer: $\ln(1 + x^2) + C$ (since $1 + x^2 > 0$, the absolute value may be dropped)

Approach:

The integrand contains $\sqrt{4 - x^2}$, so put $x = 2\sin t$ to remove the radical.

Steps:

1. Put $x = 2\sin t$, so $dx = 2\cos t\, dt$ and $\sqrt{4 - x^2} = 2\cos t$
2. $\int \sqrt{4 - x^2} dx = \int 2\cos t \cdot 2\cos t\, dt = 4\int \cos^2 t\, dt$
3. $= 4\int \frac{1 + \cos 2t}{2} dt = 2t + \sin 2t + C$
4. Substitute back: $t = \arcsin\frac{x}{2}$, and $\sin 2t = 2\sin t\cos t = \frac{x}{2}\sqrt{4 - x^2}$
5. $= 2\arcsin\frac{x}{2} + \frac{x}{2}\sqrt{4 - x^2} + C$

Answer: $2\arcsin\frac{x}{2} + \frac{x}{2}\sqrt{4 - x^2} + C$

Approach:

The derivative of the logarithm is simple, so put $u = \ln x$ and $dv = x\, dx$.

Steps:

1. Put $u = \ln x$, $dv = x dx$, so $du = \frac{1}{x} dx$ and $v = \frac{x^2}{2}$
2. $\int x \ln x dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$
3. $= \frac{x^2}{2}\ln x - \frac{1}{2}\int x dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$

Answer: $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$

Approach:

First factorise the denominator, and then split it into partial fractions.

Steps:

1. $x^2 + x - 6 = (x - 2)(x + 3)$
2. Put $\frac{2x + 3}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}$
3. Comparing after bringing to a common denominator: $2x + 3 = A(x+3) + B(x-2) = (A+B)x + (3A - 2B)$
4. Solving gives $A + B = 2$ and $3A - 2B = 3$, so $A = \frac{7}{5}$ and $B = \frac{3}{5}$
5. $\int \frac{2x+3}{x^2+x-6} dx = \frac{7}{5}\ln|x-2| + \frac{3}{5}\ln|x+3| + C$

Answer: $\frac{7}{5}\ln|x-2| + \frac{3}{5}\ln|x+3| + C$

Approach:

Reduce the power with the half-angle formula and then integrate.

Steps:

1. $\cos^2 x = \frac{1 + \cos 2x}{2}$
2. $\int \cos^2 x dx = \int \frac{1 + \cos 2x}{2} dx$
3. $= \frac{x}{2} + \frac{\sin 2x}{4} + C$

Answer: $\frac{x}{2} + \frac{\sin 2x}{4} + C$

Approach:

First split it up using linearity, and then evaluate it with the Newton-Leibniz formula.

Steps:

1. $\int_0^2 (x^3 + 1) dx = [\frac{x^4}{4} + x]_0^2$
2. $= (\frac{16}{4} + 2) - (0 + 0) = 4 + 2 = 6$

Answer: $6$

Approach:

Both $x^5$ and $x^3$ are odd functions, and the integral of an odd function over a symmetric interval is zero.

Steps:

1. By the property of odd functions: $\int_{-1}^{1} x^5 dx = 0$ and $\int_{-1}^{1} x^3 dx = 0$
2. $\int_{-1}^{1} (x^5 + x^3 + 2) dx = 0 + 0 + \int_{-1}^{1} 2 dx = 2 \cdot 2 = 4$

Answer: $4$

Approach:

The upper limit is $x^2$, so use the formula for differentiating an integral with variable limits together with the chain rule.

Steps:

1. $F'(x) = e^{x^2} \sin(x^2) \cdot \frac{d}{dx}(x^2)$
2. $= e^{x^2} \sin(x^2) \cdot 2x = 2x e^{x^2} \sin(x^2)$

Answer: $2x e^{x^2} \sin(x^2)$

Approach:

This is an improper integral of the first kind; it converges when $p > 1$, and you compute it from the definition.

Steps:

1. $\int_1^{+\infty} \frac{1}{x^2} dx = \lim_{b \to +\infty} \int_1^b x^{-2} dx$
2. $= \lim_{b \to +\infty} [-\frac{1}{x}]_1^b = \lim_{b \to +\infty} (-\frac{1}{b} + 1) = 1$
3. The limit exists, so the integral converges

Answer: It converges, and its value is $1$

Approach:

For the area use $\int_a^b f(x) dx$, and for the volume of the solid of revolution use $V = \pi\int_a^b [f(x)]^2 dx$.

Steps:

1. Area: $S = \int_0^1 x^2 dx = [\frac{x^3}{3}]_0^1 = \frac{1}{3}$
2. Volume: $V = \pi \int_0^1 (x^2)^2 dx = \pi \int_0^1 x^4 dx = \pi [\frac{x^5}{5}]_0^1 = \frac{\pi}{5}$

Answer: The area is $\frac{1}{3}$ and the volume is $\frac{\pi}{5}$

Approach:

The spring force is $F(x) = kx$, and the work is the integral of the force with respect to the displacement.

Steps:

1. $W = \int_0^a kx \, dx$
2. $= k[\frac{x^2}{2}]_0^a = \frac{ka^2}{2}$

Answer: $\frac{ka^2}{2}$