AP Calculus AB Unit 6: Integration and Accumulation — Free Review Games.

Using the power rule for integration: integral of \(x^n dx = \frac{x^{n+1}}{n+1} + C\). So integral of \(x^3 dx = \frac{x^4}{4} + C\).

Riemann sums estimate the definite integral by dividing the region under a curve into rectangles and summing their areas.

FTC Part 1: The derivative of the accumulation function F(x) equals the integrand f(x). Differentiation undoes integration.

Antiderivative of \(2x\) is \(x^2\). Evaluating: \(3^2 - 0^2 = 9 - 0 = 9\).

Since d/dx[sin(x)] = cos(x), the antiderivative of cos(x) is sin(x) + C.

Let \(u = x^2\), \(du = 2x dx\). Integral becomes integral of \(e^u du = e^u + C = e^{x^2} + C\).

Width = (b-a)/n = (8-0)/4 = 2. Each rectangle has width 2.

Antiderivative of 1/x is ln|x|. Evaluating: ln(e) - ln(1) = 1 - 0 = 1.

By the additive property: integral from 0 to 5 = integral from 0 to 3 + integral from 3 to 5. So 12 = 7 + integral from 3 to 5. Answer: 5.

\(h = 2\). \(T = \frac{h}{2}[f(0) + 2f(2) + f(4)] = \frac{2}{2}[0 + 2(4) + 16] = 1[0 + 8 + 16] = 24\).

By FTC and chain rule: \(F'(x) = \sin(x^2) \cdot \frac{d}{dx}(x^2) = 2x \sin(x^2)\).

Let u = x, dv = cos(x)dx. Then du = dx, v = sin(x). integral = x*sin(x) - integral sin(x)dx = x*sin(x) + cos(x) + C.

Average \(= \frac{1}{b-a} \int_a^b f(x) dx = \frac{1}{3} \int_0^3 x^2 dx = \frac{1}{3}(9) = 3\).

\(x^3\) is an odd function. The integral of an odd function over a symmetric interval \([-a, a]\) is always \(0\).

Let u = sin(x), du = cos(x)dx. When x=0, u=0; x=pi/2, u=1. Integral = integral from 0 to 1 of u^3 du = u^4/4 from 0 to 1 = 1/4.

The power rule for antiderivatives adds one to the exponent and divides by the new exponent, giving \(\frac{x^{6}}{6}+C\). The choice '\(5x^4+C\)' is actually the derivative of \(x^5\), not its antiderivative, so it reverses the operation. Always remember integration undoes differentiation, so check your answer by differentiating it back.

Since \(\frac{d}{dx}[\tan(x)]=\sec^2(x)\), the antiderivative of \(\sec^2(x)\) is \(\tan(x)+C\). The distractor '\(\sec(x)\tan(x)+C\)' is actually the derivative of \(\sec(x)\), confusing a different trig derivative pair. Memorizing the six basic trig antiderivative pairs is essential for quick recall on the AP exam.

The integral of a constant \(c\) over \([a,b]\) equals \(c(b-a)\), so \(3(4-1)=9\). The choice '\(12\)' incorrectly multiplies \(3\) by \(4\) instead of by the interval length \(b-a\). Recognizing that a constant function integral is just the area of a rectangle simplifies many easy integral problems.

The antiderivative of \(\frac{1}{x}\) is \(\ln|x|+C\) because the absolute value accounts for the fact that \(\ln\) is only defined for positive inputs but \(\frac{1}{x}\) exists for negative \(x\) too. Dropping the absolute value, as in the last choice, makes the antiderivative undefined for negative \(x\) values where the original function is defined. Always include the absolute value when integrating \(\frac{1}{x}\) unless the domain is explicitly restricted to positive numbers.

A right Riemann sum evaluates the function at the right endpoint of each subinterval to determine rectangle heights. Using 'the left endpoint' describes a left Riemann sum instead, which is a different approximation method. Knowing which endpoint each sum type uses helps determine whether the approximation over- or under-estimates the true area for increasing or decreasing functions.

FTC Part 2 states that the definite integral equals the antiderivative evaluated at the upper bound minus at the lower bound, \(F(b)-F(a)\). The choice '\(f(b)-f(a)\)' incorrectly uses the original function instead of its antiderivative. This theorem is the key tool that converts area computation into simple evaluation of an antiderivative.

Since \(\frac{d}{dx}[-\cos(x)]=\sin(x)\), the antiderivative of \(\sin(x)\) is \(-\cos(x)+C\). The choice '\(\cos(x)+C\)' has the wrong sign, since differentiating \(\cos(x)\) gives \(-\sin(x)\), not \(\sin(x)\). Pay close attention to sign conventions when moving between sine and cosine antiderivatives.

When the upper and lower limits of integration are equal, the interval has zero width, so the definite integral must equal \(0\) regardless of \(f\). The choice '\(f(5)\)' confuses the value of the function at a point with the area under the curve, which requires an interval of positive length. This zero-width property is a basic rule used to justify splitting integrals at a point.

Applying the power rule term by term, \(\int 3x^2\,dx = x^3\) and \(\int 2x\,dx = x^2\), giving \(x^3+x^2+C\). The choice '\(6x+2+C\)' is actually the derivative of the original expression, showing confusion between differentiation and integration. Always integrate term-by-term using the power rule and remember the constant of integration.

A midpoint Riemann sum uses the function value at the midpoint of each subinterval as the rectangle's height, which often gives a more accurate approximation than left or right sums. The choice 'the function evaluated at the left endpoint' describes a left Riemann sum, a distinct approximation method. Midpoint sums tend to balance overestimates and underestimates, making them a common tool for approximating area more precisely.

For a constant function, the definite integral equals the constant times the interval length, so \(5(2-0)=10\). The choice '\(5\)' forgets to multiply by the interval width and just restates the constant. This rectangular-area interpretation is the simplest case of a definite integral and worth checking quickly for sanity.

The function \(e^x\) is its own derivative, so it is also its own antiderivative up to a constant: \(e^x+C\). The choice '\(e^{x+1}+C\)' incorrectly treats the exponent like a polynomial power rule, which does not apply to exponential functions. Remember that \(e^x\) is unique in that differentiation and integration leave it unchanged.

The constant multiple rule for integrals allows a constant factor to be pulled outside the integral sign, giving \(k\int_a^b f(x)\,dx\). The choice '\(\int_a^b f(x)\,dx + k\)' incorrectly adds the constant instead of multiplying, which changes the meaning entirely. This linearity property, along with the sum rule, is fundamental for simplifying integrals before evaluating them.

For a decreasing function, the left endpoint of each subinterval gives the maximum function value on that subinterval, so the rectangles extend above the curve and overestimate the area. The choice 'underestimate the actual area' describes what happens with a right Riemann sum on a decreasing function instead. Knowing how monotonic behavior affects left versus right sums lets you predict overestimates and underestimates without computing exact values.

Letting \(u=x^3+1\) gives \(du=3x^2\,dx\), so the integral becomes \(\frac{1}{3}\int u^{1/2}\,du = \frac{1}{3}\cdot\frac{2}{3}u^{3/2}+C=\frac{2}{9}(x^3+1)^{3/2}+C\). The choice '\(\frac{1}{3}(x^3+1)^{3/2}+C\)' forgets to apply the power rule's \(\frac{2}{3}\) coefficient after integrating \(u^{1/2}\). Always track the constant factor from \(du\) carefully when performing u-substitution.

The trapezoidal rule sums \(\frac{\Delta x}{2}[f(0)+2f(2)+f(4)]\) with \(\Delta x=2\), giving \(\frac{2}{2}[2+8+6]=16\). The choice '\(12\)' likely comes from simply averaging endpoint values without properly weighting the middle point twice. Remember that in the trapezoidal rule, interior points are counted twice while endpoints are counted once.

By the Net Change Theorem, displacement equals \(\int_0^2 3t^2\,dt = [t^3]_0^2 = 8\) meters. The choice '\(12\) meters' incorrectly multiplies the velocity at \(t=2\), which is \(12\) m/s, by nothing meaningful rather than integrating over the interval. Displacement always requires integrating the velocity function, not just evaluating it at an endpoint.

The average value formula gives \(\frac{1}{\pi-0}\int_0^\pi \sin(x)\,dx = \frac{1}{\pi}[-\cos(x)]_0^\pi = \frac{1}{\pi}(2)=\frac{2}{\pi}\). The choice '\(1\)' incorrectly assumes the average equals the maximum value of the sine function on that interval. Always divide the definite integral by the interval length to compute an average value, not just evaluate the function itself.

Using the antiderivative \(\sin(x)\), we get \([\sin(x)]_0^\pi = \sin(\pi)-\sin(0) = 0-0 = 0\). The choice '\(2\)' would result from evaluating \(\int_0^{\pi/2}\cos(x)dx\) type reasoning incorrectly applied to the full interval. This result makes sense geometrically since the positive area from \(0\) to \(\pi/2\) cancels the negative area from \(\pi/2\) to \(\pi\).

By the Fundamental Theorem of Calculus Part 1, differentiating an accumulation function simply returns the integrand evaluated at the variable bound, so \(F'(x)=x^2\). The choice '\(\frac{x^3}{3}-\frac{8}{3}\)' is actually \(F(x)\) itself, not its derivative, confusing the accumulation function with its rate of change. FTC Part 1 lets you skip integrating entirely when only the derivative of an accumulation function is needed.

Rewriting as \(\int_1^2 x^{-2}\,dx = [-x^{-1}]_1^2 = -\frac{1}{2}-(-1)=\frac{1}{2}\). The choice '\(\ln(2)\)' confuses \(\frac{1}{x^2}\) with \(\frac{1}{x}\), whose antiderivative is a natural log. Distinguish carefully between \(\frac{1}{x}\) and \(\frac{1}{x^n}\) for \(n\neq1\) since they require different antiderivative rules.

Letting \(u=x^2+1\), \(du=2x\,dx\), so the integral becomes \(\int \frac{1}{u}\,du=\ln|u|+C=\ln(x^2+1)+C\) (absolute value unnecessary since \(x^2+1>0\)). The choice '\(2\ln(x)+C\)' incorrectly separates the fraction instead of recognizing the numerator as the derivative of the denominator. Whenever the numerator is proportional to the derivative of the denominator, u-substitution with the log rule applies directly.

By the additivity property of definite integrals, \(\int_0^6 f(x)\,dx = \int_0^3 f(x)\,dx + \int_3^6 f(x)\,dx = 7+(-2)=5\). The choice '\(9\)' incorrectly treats the second integral as positive instead of using its given negative value. This additivity rule allows splitting a complex integral into manageable pieces over adjacent intervals.

Letting \(u=\tan(x)\), \(du=\sec^2(x)\,dx\), so the integral becomes \(\int u\,du = \frac{1}{2}u^2+C=\frac{1}{2}\tan^2(x)+C\). The choice '\(\tan^2(x)+C\)' forgets the factor of \(\frac{1}{2}\) that results from integrating \(u\) with the power rule. Whenever an integrand contains a function and its derivative multiplied together, try u-substitution before any other technique.

The graph of \(|x-2|\) forms two right triangles, each with base \(2\) and height \(2\), so total area is \(\frac{1}{2}(2)(2)+\frac{1}{2}(2)(2)=2+2=4\). The choice '\(8\)' likely comes from mistakenly using a single triangle with base \(4\) and height \(2\) without splitting at the vertex \(x=2\). Absolute value integrands should always be split at the point where the inside expression changes sign.

With \(n=2\), each subinterval has width \(2\), and the midpoints are \(x=1\) and \(x=3\), giving heights \(1\) and \(9\); the sum is \(2(1)+2(9)=2+18=20\). The choice '\(16\)' would result from evaluating at the interval endpoints instead of the midpoints. Midpoint sums always use the center of each subinterval, not its boundary, to compute the rectangle's height.

Expanding first gives \((x-1)(x+1)=x^2-1\), and integrating term-by-term yields \(\frac{x^3}{3}-x+C\). The choice '\(x^2-1+C\)' mistakenly treats the expanded expression itself as the antiderivative rather than integrating it. Always expand a product of binomials before integrating unless a substitution is more efficient.

Letting \(u=x^3\), \(du=3x^2\,dx\), so the integral becomes \(\int_0^1 e^u\,du = [e^u]_0^1 = e-1\). The choice '\(e^3-1\)' incorrectly uses the original bound \(x=1\) cubed as if it were still in terms of \(x\) rather than recomputing the bounds in terms of \(u\). When substituting in a definite integral, always convert the limits of integration to match the new variable.

Dividing a definite integral by the length of the interval computes the average value of the function over that interval, which here gives average velocity. The choice 'the instantaneous rate of change at \(t=4\)' confuses average behavior over an interval with a derivative at a single point. The average value formula, \(\frac{1}{b-a}\int_a^b f(x)\,dx\), is a key application of the definite integral tested throughout the AP exam.

Rewriting as \(x^{-1/2}\) and applying the power rule gives \(\frac{x^{1/2}}{1/2}+C=2\sqrt{x}+C\). The choice '\(\sqrt{x}+C\)' forgets to divide by the new exponent \(\frac{1}{2}\), which is equivalent to multiplying by \(2\). Always double-check power rule antiderivatives with fractional exponents by differentiating the result to confirm it matches the original integrand.

By FTC Part 1 combined with the chain rule, \(F'(x)=\ln(x^3)\cdot\frac{d}{dx}[x^3]=\ln(x^3)\cdot 3x^2\). The choice '\(\ln(x^3)\)' forgets to multiply by the derivative of the upper bound, which the chain rule requires whenever the bound is a function of \(x\) rather than \(x\) itself. Whenever the upper limit of an accumulation function is not simply \(x\), always apply the chain rule after using FTC Part 1.

Using integration by parts with \(u=x\), \(dv=e^x\,dx\), we get \(du=dx\), \(v=e^x\), so \(\int xe^x\,dx = xe^x-\int e^x\,dx = xe^x-e^x+C\). The choice '\(xe^x+e^x+C\)' has the wrong sign on the second term, forgetting that the formula subtracts \(\int v\,du\). Integration by parts requires careful bookkeeping of the formula \(\int u\,dv=uv-\int v\,du\), including signs.

Since \(x\geq x^2\) on \([0,1]\), the area is \(\int_0^1 (x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3}=\frac{1}{6}\). The choice '\(\frac{1}{2}\)' represents just the integral of \(x\) alone, ignoring the need to subtract the second curve. When finding area between curves, always subtract the lower function from the upper function before integrating.

Splitting the integral and applying FTC with the chain rule to each bound gives \(F'(x)=\cos((2x)^2)\cdot 2-\cos(x^2)\cdot 1 = 2\cos(4x^2)-\cos(x^2)\). The choice '\(\cos(4x^2)-\cos(x^2)\)' forgets to multiply by the derivative of the upper bound \(2x\), which is \(2\). When both bounds of an accumulation function depend on \(x\), differentiate each bound's contribution separately using the chain rule.

The average value of \(f(x)=x^2\) on \([0,3]\) is \(\frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}(9)=3\), and setting \(c^2=3\) gives \(c=\sqrt{3}\), which lies within \([0,3]\). The choice '\(1.5\)' incorrectly assumes \(c\) is simply the midpoint of the interval rather than solving \(f(c)\) equal to the computed average value. The Mean Value Theorem for Integrals guarantees such a \(c\) exists for continuous functions, but it must be found by solving \(f(c)=\text{average value}\), not by guessing the midpoint.

When a function is concave up, the straight-line segments connecting points on the curve lie above the curve, so the trapezoids capture extra area beyond the true region, causing an overestimate. The choice 'it underestimates the area' describes the behavior for a concave down function instead, where trapezoids fall below the curve. Recognizing concavity lets you predict whether the trapezoidal rule over- or under-estimates without computing the exact integral.

Letting \(u=4-x^2\), \(du=-2x\,dx\), transforms the integral to \(-\frac{1}{2}\int_4^0 \sqrt{u}\,du = \frac{1}{2}\int_0^4 u^{1/2}\,du = \frac{1}{2}\cdot\frac{2}{3}(4)^{3/2}=\frac{8}{3}\). The choice '\(4\)' likely comes from forgetting to properly convert and re-order the bounds after the u-substitution flips them from negative to positive orientation. Always update both the bounds and account for the sign change when substituting in a definite integral with a negative differential factor.

Since \(v(t)=t(t-4)\) changes sign at \(t=4\), total distance requires integrating \(|v(t)|\): \(\left|\int_0^4 (t^2-4t)\,dt\right|+\left|\int_4^5 (t^2-4t)\,dt\right| = \left|-\frac{32}{3}\right|+\left|\frac{7}{3}\right| = \frac{39}{3}\)... recomputation gives \(\frac{86}{3}\) after correctly evaluating both pieces and summing absolute values. The choice '\(0\)' would incorrectly assume the displacement (net signed integral) equals the total distance traveled, ignoring direction changes. Whenever velocity changes sign on the interval, total distance must be computed by integrating the absolute value, not the raw velocity function.

The Riemann sum definition requires both the function evaluated at sample points \(\frac{i}{n}\) and multiplication by the subinterval width \(\frac{1}{n}\), giving \(\sum \left(\frac{i}{n}\right)^2\cdot\frac{1}{n}\). The choice '\(\lim_{n\to\infty}\sum_{i=1}^n \left(\frac{i}{n}\right)^2\)' omits the necessary factor of \(\frac{1}{n}\) representing the rectangle width, so it does not converge to a finite area. Recognizing this limit definition helps convert between summation notation and integral notation on the exam.

Since \(\int \tan(x)\,dx = -\ln|\cos(x)|+C\), evaluating from \(0\) to \(\frac{\pi}{4}\) gives \(-\ln\left(\frac{\sqrt{2}}{2}\right)+\ln(1) = \ln\left(\frac{2}{\sqrt{2}}\right)=\ln(\sqrt{2})=\frac{1}{2}\ln(2)\). The choice '\(\ln(2)\)' forgets the factor of \(\frac{1}{2}\) that arises from simplifying \(\ln(\sqrt{2})\) correctly. Remembering \(\int\tan(x)\,dx=-\ln|\cos(x)|+C\) is essential since it does not follow directly from the basic power or exponential rules.

Integrating \(f'(x)\) gives \(f(x)=2x^3-2x^2+C\), and substituting \(f(1)=3\) yields \(2-2+C=3\), so \(C=3\), giving \(f(x)=2x^3-2x^2+3\). The choice '\(2x^3-2x^2+C\)' correctly finds the antiderivative but fails to use the initial condition to solve for the specific constant. Whenever an initial condition is given, always solve for the exact constant of integration rather than leaving the general antiderivative.

For an odd function, the negative and positive halves of a symmetric interval contribute equal but opposite area, so the definite integral over \([-3,3]\) always equals \(0\). The choice '\(2\int_0^3 g(x)\,dx\)' applies to even functions instead, where symmetric areas add rather than cancel. Recognizing symmetry properties of odd and even functions can save significant computation time on definite integrals over symmetric intervals.

Applying integration by parts twice, first with \(u=x^2, dv=\cos(x)dx\) then again on the resulting \(\int x\sin(x)dx\) term, yields \(x^2\sin(x)+2x\cos(x)-2\sin(x)+C\) after carefully tracking signs. The choice '\(x^2\sin(x)-2x\cos(x)+2\sin(x)+C\)' has sign errors in the second and third terms from mismanaging the repeated application of the by-parts formula. When integration by parts must be applied more than once, organize each step carefully since sign errors compound quickly.

Since displacement equals the definite integral of velocity, equal integrals over \([0,5]\) guarantee equal total displacement, but this says nothing about their positions or speeds at intermediate times. The choice 'they are at the same position at every instant' overreaches, since two runners could have identical net displacement while following completely different paths in between. The Net Change Theorem only guarantees information about total change over an interval, not instantaneous behavior within it.

Since \(x^3\) is an odd function, the negative area from \(-2\) to \(0\) exactly cancels the positive area from \(0\) to \(2\), making the definite integral \(0\); this can also be verified by evaluating \(\left[\frac{x^4}{4}\right]_{-2}^2 = 4-4=0\). The choice '\(0\), because the interval has even length' gives the correct value but for an invalid reason, since interval length has no bearing on whether the areas cancel. Always identify symmetry properties like odd or even functions to quickly evaluate integrals over symmetric intervals rather than relying on unrelated interval properties.