Integration Basics - IB Mathematics Analysis and Approaches

Given $f(x) = \int_x^{x^2} \frac{1}{t} dt$ for $x>0$ . Find $f'(x)$ .
Answer: $\frac{1}{x}$ First, we can evaluate the integral: $f(x) = [\ln|t|]_x^{x^2}$ . Since $x>0$ , $t>0$ , so $f(x) = \ln(x^2) - \ln(x) = 2\ln(x) - \ln(x) = \ln(x)$ . Now, we find the derivative of $f(x)$ : $f'(x) = \frac{d}{dx}(\ln(x)) = \frac{1}{x}$ . Alternatively, using Leibniz rule (a more general form of the Fundamental Theorem of Calculus): If $f(x) = \int_{a(x)}^{b(x)} g(t) dt$ , then $f'(x) = g(b(x))b'(x) - g(a(x))a'(x)$ . Here, $g(t)=\frac{1}{t}$ , $b(x)=x^2$ , $a(x)=x$ . So $b'(x)=2x$ and $a'(x)=1$ . $f'(x) = \frac{1}{x^2}(2x) - \frac{1}{x}(1) = \frac{2}{x} - \frac{1}{x} = \frac{1}{x}$ .
The area of the region bounded by the curves $y=x^2+1$ and $y=x+3$ is given by which integral?
Answer: $\int_{-1}^2 (-x^2+x+2) dx$ To find the area between two curves, we first need to find their points of intersection by setting the equations equal to each other: $x^2+1 = x+3$ . This simplifies to $x^2-x-2=0$ , which factors as $(x-2)(x+1)=0$ . The points of intersection are at $x=-1$ and $x=2$ . These will be the limits of integration. To determine which function is greater on the interval $[-1, 2]$ , we can test a point, for example $x=0$ . At $x=0$ , $y=x^2+1=1$ and $y=x+3=3$ . Since $3>1$ , the line $y=x+3$ is the upper curve. The area is given by $\int_{-1}^2 (\text{upper curve} - \text{lower curve}) dx = \int_{-1}^2 ((x+3) - (x^2+1)) dx = \int_{-1}^2 (-x^2+x+2) dx$ .
Find the total area of the regions enclosed by the curve $y=x^3-4x$ and the x-axis.
Answer: 8 First, find the x-intercepts by setting $y=0$ : $x^3-4x = x(x^2-4) = x(x-2)(x+2) = 0$ . The intercepts are $x=-2, 0, 2$ . The area is split into two regions. For $x \in [-2, 0]$ , the curve is above the x-axis. For $x \in [0, 2]$ , the curve is below the x-axis. The total area is $\int_{-2}^0 (x^3-4x) dx + |\int_0^2 (x^3-4x) dx|$ . Calculate the first integral: $[\frac{x^4}{4}-2x^2]_{-2}^0 = (0) - (\frac{(-2)^4}{4}-2(-2)^2) = -(\frac{16}{4}-8) = -(4-8) = 4$ . Calculate the second integral: $[\frac{x^4}{4}-2x^2]_0^2 = (\frac{2^4}{4}-2(2)^2) - (0) = (\frac{16}{4}-8) = 4-8 = -4$ . The area for this part is $|-4|=4$ . The total area is $4+4=8$ .
The value of $\int_{-a}^a (x^3 \cos x + 1) dx$ is 6. What is the value of $a$ ?
Answer: 3 The integral can be split: $\int_{-a}^a x^3 \cos x dx + \int_{-a}^a 1 dx$ . The function $f(x) = x^3 \cos x$ is an odd function because $f(-x) = (-x)^3 \cos(-x) = -x^3 \cos x = -f(x)$ . The integral of an odd function over a symmetric interval $[-a, a]$ is 0. Therefore, the expression simplifies to $\int_{-a}^a 1 dx$ . This evaluates to $[x]_{-a}^a = a - (-a) = 2a$ . We are given that the value is 6, so $2a=6$ , which means $a=3$ .
The integral $\int_0^2 (x^2-3x+2) dx$ represents the net signed area between the curve $y=x^2-3x+2$ and the x-axis from $x=0$ to $x=2$ . What is the total area of the regions bounded by the curve and the x-axis over this interval?
Answer: 1 The curve is a parabola $y=(x-1)(x-2)$ with roots at $x=1$ and $x=2$ . Over the interval $[0, 2]$ , the curve is above the x-axis for $x \in [0, 1)$ and below the x-axis for $x \in (1, 2]$ . The total area is the sum of the absolute values of the integrals over these sub-intervals: Area = $\int_0^1 (x^2-3x+2) dx + |\int_1^2 (x^2-3x+2) dx|$ . Let $F(x) = \frac{x^3}{3} - \frac{3x^2}{2} + 2x$ . $F(0)=0$ , $F(1) = \frac{1}{3} - \frac{3}{2} + 2 = \frac{2-9+12}{6} = \frac{5}{6}$ , $F(2) = \frac{8}{3} - \frac{12}{2} + 4 = \frac{8}{3} - 6 + 4 = \frac{8}{3} - 2 = \frac{2}{3}$ . First integral: $F(1)-F(0) = \frac{5}{6}$ . Second integral: $F(2)-F(1) = \frac{2}{3} - \frac{5}{6} = \frac{4-5}{6} = -\frac{1}{6}$ . Total area = $\frac{5}{6} + |-\frac{1}{6}| = \frac{5}{6} + \frac{1}{6} = \frac{6}{6} = 1$ . The value of the single integral from 0 to 2 would be $F(2)-F(0) = 2/3$ , which is a possible distractor.
The gradient of a curve $y=f(x)$ is given by $\frac{dy}{dx} = 3x^2+k$ . The curve has a local minimum at the point $(1, -4)$ . Find the value of $f(2)$ .
Answer: 0 At a local minimum, the gradient $\frac{dy}{dx}$ is 0. So, at $x=1$ , $3(1)^2+k = 0 \Rightarrow 3+k=0 \Rightarrow k=-3$ . The gradient function is $\frac{dy}{dx} = 3x^2-3$ . To find $f(x)$ , we integrate: $f(x) = \int (3x^2-3) dx = x^3 - 3x + C$ . The curve passes through $(1, -4)$ , so we can find $C$ : $-4 = 1^3 - 3(1) + C \Rightarrow -4 = 1-3+C \Rightarrow -4 = -2+C \Rightarrow C=-2$ . So the equation of the curve is $f(x) = x^3-3x-2$ . We need to find $f(2)$ : $f(2) = 2^3 - 3(2) - 2 = 8-6-2 = 0$ .
Let $f(x)$ be a function such that $\int_1^3 f(x) dx = 5$ . Which statement must be true?
Answer: An anti-derivative $F(x)$ of $f(x)$ satisfies $F(3) - F(1) = 5$ . The statement $\int_1^3 f(x) dx = 5$ is the definition of the definite integral according to the Fundamental Theorem of Calculus, Part 2. It states that if $F$ is an anti-derivative of $f$ , then $\int_a^b f(x) dx = F(b) - F(a)$ . So, D is true by definition. A is not necessarily true, as $f(x)$ could be negative for some parts of the interval as long as the net signed area is 5. B is not necessarily true because 'area' is non-negative, whereas the definite integral can be negative. If $f(x)$ is ever negative on $[1,3]$ , the total area could be greater than 5. C is related to the Mean Value Theorem for Integrals, which states there is a $c$ such that $f(c) = \frac{1}{3-1}\int_1^3 f(x) dx = \frac{5}{2} = 2.5$ , but this theorem is only guaranteed if $f$ is continuous, which is not stated. The Fundamental Theorem of Calculus, as used in D, is the most direct and certain consequence.
The area of the region enclosed by the graph of $y = \frac{4}{2x+1}$ , the x-axis, and the lines $x=0$ and $x=k$ is $\ln(25)$ . Find the value of $k$ , where $k > 0$ .
Answer: 2 The area is given by the definite integral $\int_0^k \frac{4}{2x+1} dx$ . First, find the anti-derivative of $\frac{4}{2x+1}$ . This is $4 \cdot \frac{1}{2} \ln|2x+1| = 2\ln|2x+1|$ . Now, evaluate the definite integral: $[2\ln|2x+1|]_0^k = 2\ln(2k+1) - 2\ln(1) = 2\ln(2k+1)$ . We are given that this area is $\ln(25)$ . So, $2\ln(2k+1) = \ln(25)$ . Using logarithm properties, this becomes $\ln((2k+1)^2) = \ln(25)$ . Therefore, $(2k+1)^2 = 25$ . Since $k>0$ , we take the positive root: $2k+1 = 5$ , which gives $2k=4$ and $k=2$ .
A particle moves along a straight line with velocity $v(t) = 8-2t$ m/s for $t \ge 0$ . What is the total distance travelled by the particle in the first 6 seconds?
Answer: 20 m Total distance travelled is the integral of the speed, $|v(t)|$ . First, find when the velocity changes sign: $v(t) = 8-2t=0$ gives $t=4$ . For $0 \le t < 4$ , $v(t)>0$ . For $t > 4$ , $v(t)<0$ . The total distance is $\int_0^6 |8-2t| dt = \int_0^4 (8-2t) dt + \int_4^6 -(8-2t) dt$ . First integral: $[8t-t^2]_0^4 = (32-16)-0 = 16$ . Second integral: $\int_4^6 (2t-8) dt = [t^2-8t]_4^6 = (36-48)-(16-32) = -12 - (-16) = 4$ . Total distance = $16+4=20$ m. Distractor A (12m) is the displacement $\int_0^6 v(t) dt$ .
Evaluate $\int_{\pi/6}^{\pi/3} \sin(2x) dx$ .
Answer: $\frac{1}{2}$ First, find the anti-derivative of $\sin(2x)$ . Using the reverse chain rule, this is $-\frac{1}{2}\cos(2x)$ . Now, apply the Fundamental Theorem of Calculus: $[-\frac{1}{2}\cos(2x)]_{\pi/6}^{\pi/3} = (-\frac{1}{2}\cos(2 \cdot \frac{\pi}{3})) - (-\frac{1}{2}\cos(2 \cdot \frac{\pi}{6})) = -\frac{1}{2}\cos(\frac{2\pi}{3}) + \frac{1}{2}\cos(\frac{\pi}{3})$ . We know $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ . Substituting these values gives: $-\frac{1}{2}(-\frac{1}{2})) + \frac{1}{2}(\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ .
The region bounded by the graph of $y=\sqrt{x}$ , the x-axis, and the line $x=a$ has an area of 18. Find the value of $a$ .
Answer: 9 The area is given by the integral $\int_0^a \sqrt{x} dx = \int_0^a x^{1/2} dx$ . Evaluating the integral: $[\frac{x^{3/2}}{3/2}]_0^a = [\frac{2}{3}x^{3/2}]_0^a = \frac{2}{3}a^{3/2} - 0 = \frac{2}{3}a^{3/2}$ . We are given that this area is 18. So, $\frac{2}{3}a^{3/2} = 18$ . Multiply by $\frac{3}{2}$ : $a^{3/2} = 18 \cdot \frac{3}{2} = 27$ . To find $a$ , we can write $(a^{1/2})^3 = 3^3$ , so $a^{1/2} = 3$ . Squaring both sides gives $a=9$ .
Given that $\int_1^4 f(x) dx = 6$ and $\int_1^4 g(x) dx = -2$ , find the value of $\int_4^1 (f(x) - 2g(x)) dx$ .
Answer: -10 Using the properties of definite integrals: $\int_4^1 (f(x) - 2g(x)) dx = - \int_1^4 (f(x) - 2g(x)) dx$ . Using the linearity of integrals: $-[\int_1^4 f(x) dx - 2\int_1^4 g(x) dx])$ . Substitute the given values: $-[6 - 2(-2)]) = -[6+4] = -10$ . Distractor D (10) arises from ignoring the negative sign from swapping the limits of integration. Distractor B (-2) comes from an error like $6 - 2(2) = 2$ then flipping the sign to -2.
Let $f'(x) = 3x^2 - 2\cos(x)$ and $f(\frac{\pi}{2}) = \frac{\pi^3}{8}$ . Find $f(x)$ .
Answer: $x^3 - 2\sin(x) + 2$ To find $f(x)$ , we need to integrate $f'(x)$ : $f(x) = \int (3x^2 - 2\cos(x)) dx = x^3 - 2\sin(x) + C$ . We are given the condition $f(\frac{\pi}{2}) = \frac{\pi^3}{8}$ to find the constant of integration $C$ . Substitute $x=\frac{\pi}{2}$ : $(\frac{\pi}{2})^3 - 2\sin(\frac{\pi}{2}) + C = \frac{\pi^3}{8}$ . This simplifies to $\frac{\pi^3}{8} - 2(1) + C = \frac{\pi^3}{8}$ . So, $-2 + C = 0$ , which means $C=2$ . Therefore, $f(x) = x^3 - 2\sin(x) + 2$ .
The rate of change of a population $P$ is modelled by $\frac{dP}{dt} = k\sqrt{t}$ , where $t$ is the time in years. The initial population is $P_0$ . Find the population $P(t)$ at time $t$ .
Answer: $P_0 + \frac{2k}{3}t^{3/2}$ To find the population $P(t)$ , we need to integrate the rate of change with respect to time: $P(t) = \int k\sqrt{t} dt = \int k t^{1/2} dt$ . Using the power rule for integration, this becomes $k \frac{t^{1/2+1}}{1/2+1} + C = k \frac{t^{3/2}}{3/2} + C = \frac{2k}{3} t^{3/2} + C$ . We are given that the initial population is $P_0$ , which means at $t=0$ , $P(0) = P_0$ . Substituting this in: $P(0) = \frac{2k}{3}(0)^{3/2} + C = P_0$ , which implies $C=P_0$ . Therefore, the population at time $t$ is $P(t) = P_0 + \frac{2k}{3}t^{3/2}$ .
Let $f''(x) = 6x$ . The graph of $f(x)$ has a tangent line with equation $y=2x+3$ at the point where $x=1$ . Find $f(x)$ .
Answer: $x^3 - x + 5$ First, integrate $f''(x)=6x$ to find $f'(x)$ : $f'(x) = \int 6x dx = 3x^2 + C_1$ . The gradient of the tangent line $y=2x+3$ is 2. This means that at $x=1$ , the gradient of $f(x)$ is 2. So, $f'(1)=2$ . Substitute this into the expression for $f'(x)$ : $3(1)^2 + C_1 = 2 \Rightarrow 3+C_1=2 \Rightarrow C_1=-1$ . Thus, $f'(x) = 3x^2-1$ . Now integrate $f'(x)$ to find $f(x)$ : $f(x) = \int (3x^2-1) dx = x^3 - x + C_2$ . To find $C_2$ , we need a point on the curve $f(x)$ . At $x=1$ , the point is on the tangent line, so $y=2(1)+3=5$ . The point is $(1,5)$ . Substitute this into the expression for $f(x)$ : $f(1) = 1^3 - 1 + C_2 = 5 \Rightarrow 0+C_2=5 \Rightarrow C_2=5$ . Therefore, $f(x) = x^3 - x + 5$ .
The region R is bounded by the curve $y=e^{-x}$ , the x-axis, and the lines $x=0$ and $x=\ln(a)$ , where $a>1$ . The area of R is $\frac{2}{3}$ . Find the value of $a$ .
Answer: 3 The area of the region R is given by the integral $A = \int_0^{\ln(a)} e^{-x} dx$ . The anti-derivative of $e^{-x}$ is $-e^{-x}$ . Evaluating the definite integral: $[-e^{-x}]_0^{\ln(a)} = (-e^{-\ln(a)}) - (-e^{-0}) = -e^{\ln(a^{-1})} + e^0 = -a^{-1} + 1 = 1 - \frac{1}{a}$ . We are given that the area is $\frac{2}{3}$ . So, $1 - \frac{1}{a} = \frac{2}{3}$ . Solving for $a$ : $\frac{1}{a} = 1 - \frac{2}{3} = \frac{1}{3}$ . Therefore, $a=3$ .
Find the area of the region enclosed by the graph of $y=3x-x^2$ and the line $y=x$ .
Answer: $\frac{4}{3}$ First, find the points of intersection: $3x-x^2 = x \Rightarrow 2x-x^2 = 0 \Rightarrow x(2-x) = 0$ . The intersection points are at $x=0$ and $x=2$ . In the interval $[0, 2]$ , we test a point, e.g., $x=1$ . The parabola gives $y=3(1)-1^2=2$ and the line gives $y=1$ . So, the parabola $y=3x-x^2$ is the upper curve. The area is $A = \int_0^2 ((3x-x^2) - x) dx = \int_0^2 (2x-x^2) dx$ . Now, integrate: $[x^2 - \frac{x^3}{3}]_0^2 = (2^2 - \frac{2^3}{3}) - (0) = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3}$ .
An anti-derivative of $f(x) = (2x+3)^4$ is $F(x)$ . Which of the following could be $F(x)$ ?
Answer: $\frac{(2x+3)^5}{10}$ This requires using the reverse chain rule for integration. We are looking for $\int (2x+3)^4 dx$ . Let $u = 2x+3$ , then $du = 2 dx$ or $dx = \frac{1}{2}du$ . The integral becomes $\int u^4 (\frac{1}{2} du) = \frac{1}{2} \int u^4 du = \frac{1}{2} \frac{u^5}{5} + C = \frac{u^5}{10} + C$ . Substituting back $u=2x+3$ , we get $\frac{(2x+3)^5}{10} + C$ . Option C is a possible anti-derivative (with $C=0$ ). Distractor B is a common error where the $\frac{1}{a}$ factor from the inner function's derivative is forgotten. Distractor A is the derivative, not the anti-derivative.

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