Integration Techniques - IB Mathematics Analysis and Approaches

The gradient of a curve is given by $f'(x) = e^{x/2} - 2x$ . The curve passes through the point $(0, 5)$ . Find the equation of the curve.
Answer: $y = 2e^{x/2} - x^2 + 3$ To find the equation of the curve $y = f(x)$ , integrate the gradient function: $f(x) = \int (e^{x/2} - 2x) dx$ . Using the reverse chain rule for $e^{x/2}$ and the power rule for $-2x$ , we get $f(x) = \frac{e^{x/2}}{1/2} - \frac{2x^2}{2} + C = 2e^{x/2} - x^2 + C$ . To find $C$ , use the point $(0, 5)$ : $5 = 2e^{0/2} - 0^2 + C \implies 5 = 2e^0 + C \implies 5 = 2(1) + C \implies C = 3$ . So the equation is $y = 2e^{x/2} - x^2 + 3$ .
The area of the region enclosed by the curve $y = \frac{3}{x}$ , the x-axis, and the lines $x=1$ and $x=e^2$ is given by $\int_1^{e^2} \frac{3}{x} dx$ . Calculate this area.
Answer: 6 To find the area, we evaluate the definite integral. The antiderivative of $\frac{3}{x}$ is $3\ln|x|$ . Using the Fundamental Theorem of Calculus: $\int_1^{e^2} \frac{3}{x} dx = [3\ln|x|]_1^{e^2} = 3\ln(e^2) - 3\ln(1)$ . Using logarithm properties, $\ln(e^2) = 2$ and $\ln(1) = 0$ . So, the area is $3(2) - 3(0) = 6$ .
The value of $\int_a^b f(x) dx$ represents the geometric area of the region bounded by the curve $y=f(x)$ , the x-axis, and the lines $x=a$ and $x=b$ only if which condition is met?
Answer: $f(x) \ge 0$ for all $x$ in the interval $[a, b]$ . A definite integral $\int_a^b f(x) dx$ calculates the signed area between the curve and the x-axis. Regions below the x-axis contribute negatively to the value of the integral. For the integral's value to be exactly equal to the geometric area, the entire function must be non-negative ( $f(x) \ge 0$ ) on the interval $[a, b]$ , so that no part of the region contributes a negative value.
Let $f(x) = \sin(x^3)$ . Which of the following definite integrals is equal to zero?
Answer: $\int_{-\pi/2}^{\pi/2} f(x) dx$ The function $f(x) = \sin(x^3)$ is an odd function because $f(-x) = \sin((-x)^3) = \sin(-x^3) = -\sin(x^3) = -f(x)$ . The definite integral of an odd function over a symmetric interval of the form $[-a, a]$ is always zero. Choice A, $\int_{-\pi/2}^{\pi/2} f(x) dx$ , is an integral of an odd function over a symmetric interval, so its value is 0.
Find $\int (2x - \frac{1}{x})^2 dx$ .
Answer: $\frac{4}{3}x^3 - 4x - \frac{1}{x} + C$ First, expand the integrand: $(2x - \frac{1}{x})^2 = (2x)^2 - 2(2x)(\frac{1}{x}) + (\frac{1}{x})^2 = 4x^2 - 4 + \frac{1}{x^2} = 4x^2 - 4 + x^{-2}$ . Now, integrate this expression term by term: $\int (4x^2 - 4 + x^{-2}) dx = \frac{4x^3}{3} - 4x + \frac{x^{-1}}{-1} + C = \frac{4}{3}x^3 - 4x - \frac{1}{x} + C$ .
Find $\int \frac{3x^2+2}{x^2} dx$ .
Answer: $3x - \frac{2}{x} + C$ Before integrating, simplify the fraction by splitting it into two terms: $\frac{3x^2+2}{x^2} = \frac{3x^2}{x^2} + \frac{2}{x^2} = 3 + 2x^{-2}$ . Now, integrate term by term: $\int (3 + 2x^{-2}) dx = 3x + \frac{2x^{-1}}{-1} + C = 3x - 2x^{-1} + C = 3x - \frac{2}{x} + C$ .
A particle moves along a straight line with velocity $v(t) = 6t^2 - 4t + 1$ m/s. Find the displacement of the particle between $t=1$ and $t=3$ .
Answer: 38 m Displacement is the definite integral of velocity over the time interval. We need to calculate $\int_1^3 (6t^2 - 4t + 1) dt$ . First, find the antiderivative: $\int (6t^2 - 4t + 1) dt = \frac{6t^3}{3} - \frac{4t^2}{2} + t = 2t^3 - 2t^2 + t$ . Now evaluate this from $t=1$ to $t=3$ : $[2t^3 - 2t^2 + t]_1^3 = (2(3)^3 - 2(3)^2 + 3) - (2(1)^3 - 2(1)^2 + 1) = (2(27) - 2(9) + 3) - (2 - 2 + 1) = (54 - 18 + 3) - (1) = 39 - 1 = 38$ . The displacement is 38 m.
Given that $\int_0^k (2x - 3) dx = 4$ , and $k > 0$ , find the value of $k$ .
Answer: 4 First, evaluate the definite integral in terms of $k$ . The antiderivative of $2x-3$ is $x^2 - 3x$ . Evaluate this from 0 to $k$ : $[x^2 - 3x]_0^k = (k^2 - 3k) - (0^2 - 3(0)) = k^2 - 3k$ . Now set this expression equal to 4: $k^2 - 3k = 4$ . Rearrange into a standard quadratic equation: $k^2 - 3k - 4 = 0$ . Factor the quadratic: $(k-4)(k+1) = 0$ . The possible values for $k$ are $k=4$ and $k=-1$ . Since the problem states that $k > 0$ , the correct answer is $k=4$ .
Find the indefinite integral $\int e^{3-2x} dx$ .
Answer: $-\frac{1}{2}e^{3-2x} + C$ This integral requires the reverse chain rule for a function of the form $\int f(ax+b) dx = \frac{1}{a}F(ax+b) + C$ . Here, $f(u)=e^u$ , and the inner function is $3-2x$ . The antiderivative of $e^u$ is $e^u$ . The derivative of the inner function is -2. Therefore, we must divide by -2. $\int e^{3-2x} dx = \frac{1}{-2}e^{3-2x} + C = -\frac{1}{2}e^{3-2x} + C$ .
Given that $\int_1^5 f(x) dx = 8$ and $\int_1^3 f(x) dx = 5$ , what is the value of $\int_3^5 2f(x) dx$ ?
Answer: 6 Using the property of definite integrals, $\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx$ . We have $\int_1^5 f(x) dx = \int_1^3 f(x) dx + \int_3^5 f(x) dx$ . Substituting the given values: $8 = 5 + \int_3^5 f(x) dx$ . This implies that $\int_3^5 f(x) dx = 8 - 5 = 3$ . The question asks for $\int_3^5 2f(x) dx$ , which is equal to $2\int_3^5 f(x) dx = 2(3) = 6$ .
Find $\int \cos(4x - \frac{\pi}{2}) dx$ .
Answer: $\frac{1}{4}\sin(4x - \frac{\pi}{2}) + C$ This requires the reverse chain rule. The integral of $\cos(u)$ is $\sin(u)$ . The inner function is $u = 4x - \frac{\pi}{2}$ , and its derivative is $u' = 4$ . We must divide the result by this derivative. Thus, $\int \cos(4x - \frac{\pi}{2}) dx = \frac{1}{4}\sin(4x - \frac{\pi}{2}) + C$ .
Given that the derivative of a function is $f'(x) = 3x^2 - \frac{4}{x^2}$ and the function passes through the point $(2, 9)$ , find $f(x)$ .
Answer: $f(x) = x^3 + \frac{4}{x} - 1$ To find $f(x)$ , we must integrate $f'(x)$ . First, rewrite $f'(x)$ as $3x^2 - 4x^{-2}$ . The integral is $f(x) = \int (3x^2 - 4x^{-2}) dx = \frac{3x^3}{3} - \frac{4x^{-1}}{-1} + C = x^3 + 4x^{-1} + C = x^3 + \frac{4}{x} + C$ . Now use the given point $(2, 9)$ to solve for the constant of integration, $C$ . Substitute $x=2$ and $f(x)=9$ : $9 = 2^3 + \frac{4}{2} + C \implies 9 = 8 + 2 + C \implies 9 = 10 + C \implies C = -1$ . Therefore, the function is $f(x) = x^3 + \frac{4}{x} - 1$ .
Find $\int \frac{(x-2)^2}{\sqrt{x}} dx$ .
Answer: $\frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8\sqrt{x} + C$ First, expand the numerator and rewrite the expression with fractional exponents: $\frac{(x-2)^2}{\sqrt{x}} = \frac{x^2 - 4x + 4}{x^{1/2}}$ . Then, divide each term in the numerator by $x^{1/2}$ : $x^{3/2} - 4x^{1/2} + 4x^{-1/2}$ . Now, integrate term by term using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ : $\int (x^{3/2} - 4x^{1/2} + 4x^{-1/2}) dx = \frac{x^{5/2}}{5/2} - \frac{4x^{3/2}}{3/2} + \frac{4x^{1/2}}{1/2} + C$ . Simplifying gives: $\frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8x^{1/2} + C = \frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + 8\sqrt{x} + C$ .
Evaluate the definite integral $\int_0^{\pi/2} 3\sin(2x) dx$ .
Answer: 3 First, find the antiderivative of $3\sin(2x)$ . Using the reverse chain rule, $\int 3\sin(2x) dx = 3 \cdot \frac{-\cos(2x)}{2} + C = -\frac{3}{2}\cos(2x) + C$ . Now evaluate the definite integral: $[-\frac{3}{2}\cos(2x)]_0^{\pi/2} = (-\frac{3}{2}\cos(2 \cdot \frac{\pi}{2})) - (-\frac{3}{2}\cos(2 \cdot 0)) = (-\frac{3}{2}\cos(\pi)) - (-\frac{3}{2}\cos(0))$ . Since $\cos(\pi) = -1$ and $\cos(0) = 1$ , this becomes $(-\frac{3}{2}(-1)) - (-\frac{3}{2}(1)) = \frac{3}{2} - (-\frac{3}{2}) = \frac{3}{2} + \frac{3}{2} = 3$ .
Find $\int \frac{5}{\sqrt{3x+4}} dx$ .
Answer: $\frac{10}{3}\sqrt{3x+4} + C$ First, rewrite the integrand with a negative fractional exponent: $\int 5(3x+4)^{-1/2} dx$ . Apply the reverse chain rule. The antiderivative of $u^{-1/2}$ is $\frac{u^{1/2}}{1/2} = 2u^{1/2}$ . The inner function is $3x+4$ , and its derivative is 3. We must divide by 3. So, the integral is $5 \cdot \frac{(3x+4)^{1/2}}{1/2} \cdot \frac{1}{3} + C = 5 \cdot 2\sqrt{3x+4} \cdot \frac{1}{3} + C = \frac{10}{3}\sqrt{3x+4} + C$ .
Find the exact value of $\int_{\pi/6}^{\pi/3} (2\cos x - \sin x) dx$ .
Answer: $\frac{\sqrt{3}-1}{2}$ First, find the antiderivative of $2\cos x - \sin x$ . $\int (2\cos x - \sin x) dx = 2\sin x - (-\cos x) = 2\sin x + \cos x$ . Now, apply the Fundamental Theorem of Calculus: $[2\sin x + \cos x]_{\pi/6}^{\pi/3} = (2\sin(\frac{\pi}{3}) + \cos(\frac{\pi}{3})) - (2\sin(\frac{\pi}{6}) + \cos(\frac{\pi}{6}))$ . Using exact values: $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}, \cos(\frac{\pi}{3})=\frac{1}{2}, \sin(\frac{\pi}{6})=\frac{1}{2}, \cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$ . Substitution gives: $(2(\frac{\sqrt{3}}{2}) + \frac{1}{2}) - (2(\frac{1}{2}) + \frac{\sqrt{3}}{2}) = (\sqrt{3} + \frac{1}{2}) - (1 + \frac{\sqrt{3}}{2}) = \sqrt{3} + \frac{1}{2} - 1 - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2}$ .
The rate of change of a quantity $P$ is modelled by the equation $\frac{dP}{dt} = \frac{10}{2t+1}$ . Given that $P=15$ when $t=0$ , find the value of $P$ when $t=1$ .
Answer: $15 + 5\ln(3)$ To find $P(t)$ , we integrate the rate of change: $P(t) = \int \frac{10}{2t+1} dt$ . Using the reverse chain rule for $\frac{1}{ax+b}$ , the integral is $10 \cdot \frac{1}{2} \ln|2t+1| + C = 5\ln|2t+1| + C$ . We are given that $P=15$ when $t=0$ . So, $15 = 5\ln|2(0)+1| + C = 5\ln(1) + C = 5(0) + C \implies C=15$ . The equation for $P$ is $P(t) = 5\ln|2t+1| + 15$ . Now, we find $P$ when $t=1$ : $P(1) = 5\ln|2(1)+1| + 15 = 5\ln(3) + 15$ .
Find $\int \frac{6}{\sqrt[3]{x^2}} dx$ .
Answer: $18\sqrt[3]{x} + C$ First, rewrite the integrand using a negative fractional exponent: $\frac{6}{\sqrt[3]{x^2}} = \frac{6}{x^{2/3}} = 6x^{-2/3}$ . Now apply the power rule for integration: $\int 6x^{-2/3} dx = 6 \cdot \frac{x^{-2/3 + 1}}{-2/3 + 1} + C = 6 \cdot \frac{x^{1/3}}{1/3} + C$ . Dividing by $1/3$ is the same as multiplying by 3, so this simplifies to $6 \cdot 3x^{1/3} + C = 18x^{1/3} + C = 18\sqrt[3]{x} + C$ .
The area of the region bounded by the curve $y = x^2 - 4$ , the x-axis, from $x=0$ to $x=2$ is given by which expression?
Answer: $\int_0^2 (4-x^2) dx$ The function $y = x^2 - 4$ is non-positive on the interval $[0, 2]$ . For example, at $x=1$ , $y = 1-4 = -3$ . To find the geometric area, we must integrate the absolute value of the function, $|x^2 - 4|$ . Since $x^2 - 4 \le 0$ on $[0, 2]$ , this is equivalent to integrating $-(x^2 - 4)$ , which is $4 - x^2$ . Therefore, the correct expression for the area is $\int_0^2 (4-x^2) dx$ . Option A would give a negative result, which is the signed area, not the geometric area.
If $\int f(x) dx = g(x) + C$ , where $C$ is the constant of integration, which of the following statements must be true?
Answer: $g'(x) = f(x)$ The definition of an indefinite integral is that it is the antiderivative of the integrand. This means that if you differentiate the result of the integration, you get back the original function. Therefore, if $\int f(x) dx = g(x) + C$ , then $\frac{d}{dx}(g(x) + C) = f(x)$ . Since the derivative of a constant $C$ is zero, this simplifies to $g'(x) = f(x)$ .

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