When you see ∫f′(x) f(x) dx∫f′(x)f(x)dx, which method is usually simplest to start with?
A By parts method
B Partial fractions
C Trig identities
D Substitution method
If a function and its derivative appear together, choose u=f(x)u=f(x). Then du=f′(x) dxdu=f′(x)dx, turning the integral into ∫u du∫udu, which is a basic standard form.
Which derivative–integral pair is correct for basic calculus practice?
A ∫cosx∫cosx → −sinx−sinx
B ∫sinx∫sinx → cosxcosx
C ∫cosx∫cosx → sinxsinx
D ∫sinx∫sinx → tanxtanx
Since ddx(sinx)=cosxdxd(sinx)=cosx, the antiderivative of cosxcosx is sinx+Csinx+C. This is one of the most-used standard integrals.
A good first choice for ∫xlnx dx∫xlnxdx is which method?
A By parts method
B Substitution method
C Partial fractions
D Symmetry property
The integrand is a product of xx and lnxlnx. By parts with u=lnxu=lnx, dv=x dxdv=xdx simplifies it into polynomial terms that are easy to integrate.
Which integral equals 1aln∣ax+b∣+Ca1ln∣ax+b∣+C for a≠0a=0?
A ∫(ax+b) dx∫(ax+b)dx
B ∫1ax+b dx∫ax+b1dx
C ∫1×2 dx∫x21dx
D ∫ln(ax+b) dx∫ln(ax+b)dx
Let u=ax+bu=ax+b. Then du=a dxdu=adx, so the integral becomes 1a∫1u du=1aln∣u∣+Ca1∫u1du=a1ln∣u∣+C.
Which method is most suitable for ∫x+1×2+x dx∫x2+xx+1dx?
A By parts method
B Trig substitution
C Partial fractions
D Numerical method
This is a rational function. Factor the denominator x2+x=x(x+1)x2+x=x(x+1). The expression simplifies, and partial fractions (or cancellation) leads to standard ln∣x∣ln∣x∣ type integrals.
What is the main geometric meaning of ∫abf(x) dx∫abf(x)dx (basic)?
A Signed area
B Tangent slope
C Maximum value
D Curve length
The definite integral measures net area: parts above the xx-axis count positive and parts below count negative. For total geometric area, split intervals and use absolute values.
Which statement about definite integrals is always true?
A ∫abf=f(b)∫abf=f(b)
B ∫abf=b−a∫abf=b−a
C ∫aaf=0∫aaf=0
D ∫abf≥0∫abf≥0
Over an interval of zero length, there is no accumulation. So the definite integral from a point to the same point is always zero for any integrable function.
If limits are reversed, ∫baf(x) dx∫baf(x)dx becomes
A Negative of original
B Same as original
C Square of original
D Double of original
Reversing limits changes direction of accumulation, so ∫abf(x) dx=−∫baf(x) dx∫abf(x)dx=−∫baf(x)dx. This property is used to simplify expressions and compare integrals.
Which formula correctly gives average value of ff on [a,b][a,b]?
A ∫abf∫abf
B 1b−a∫abfb−a1∫abf
C f(a)+f(b)22f(a)+f(b)
D ∫abf′∫abf′
Average value equals total accumulation divided by interval length. It matches a constant height that produces the same area under the curve across [a,b][a,b].
FTC Part II is mainly used to
A Factor polynomials
B Solve triangles
C Evaluate definite integral
D Find asymptotes
FTC Part II states: if F′(x)=f(x)F′(x)=f(x), then ∫abf(x) dx=F(b)−F(a)∫abf(x)dx=F(b)−F(a). It turns definite integrals into straightforward substitution.
For G(x)=∫1xt3 dtG(x)=∫1xt3dt, what is G′(x)G′(x)?
A x3x3
B 3x23x2
C x4x4
D t3t3
By FTC Part I, differentiating an accumulation function gives the integrand evaluated at the upper limit. So G′(x)=x3G′(x)=x3, assuming continuity.
Which identity helps integrate cos2xcos2x easily?
A Double-angle only
B Product-to-sum
C Limit definition
D Half-angle identity
Use cos2x=1+cos2x2cos2x=21+cos2x. This converts a square into a sum of simple terms, making integration immediate.
Which integral is best handled using trig substitution?
A ∫ex dx∫exdx
B ∫x2 dx∫x2dx
C ∫a2−x2 dx∫a2−x2dx
D ∫1x dx∫x1dx
Expressions like a2−x2a2−x2 simplify with x=asinθx=asinθ. Then the root becomes acosθacosθ, reducing the integral to basic trig forms.
A standard result is ∫sec2x dx∫sec2xdx equals
A secx+Csecx+C
B tanx+Ctanx+C
C −cotx+C−cotx+C
D sinx+Csinx+C
Since ddx(tanx)=sec2xdxd(tanx)=sec2x, the integral of sec2xsec2x is tanx+Ctanx+C. This is used often in substitution.
Which function gives −ln∣x∣+C−ln∣x∣+C after integration?
A ∫−1x dx∫x−1dx
B ∫1×2 dx∫x21dx
C ∫lnx dx∫lnxdx
D ∫x−3 dx∫x−3dx
Because ∫1x dx=ln∣x∣+C∫x1dx=ln∣x∣+C, multiplying by −1−1 gives −ln∣x∣+C−ln∣x∣+C. The absolute value ensures correctness for negative xx.
Which definite integral must be treated as improper (intro)?
A ∫01x dx∫01xdx
B ∫12×2 dx∫12x2dx
C ∫011x dx∫01x1dx
D ∫−11x dx∫−11xdx
The integrand 1xx1 becomes infinite at x=0x=0. Such integrals are defined using limits and may converge to a finite value.
For an odd function ff, ∫−aaf(x) dx∫−aaf(x)dx equals
A 00
B 2∫0af2∫0af
C ∫0af∫0af
D Always positive
Odd functions satisfy f(−x)=−f(x)f(−x)=−f(x). Over symmetric limits, positive and negative areas cancel exactly, so the net signed area is zero.
For an even function ff, ∫−aaf(x) dx∫−aaf(x)dx equals
A 00
B ∫0af∫0af
C 2∫0af2∫0af
D −2∫0af−2∫0af
Even functions satisfy f(−x)=f(x)f(−x)=f(x). The area on [−a,0][−a,0] equals the area on [0,a][0,a], so the total doubles the half-interval integral.
Area between y=f(x)y=f(x) and y=g(x)y=g(x) needs first
A Second derivative
B Intersection points
C Taylor expansion
D Log rules
To set correct limits, solve f(x)=g(x)f(x)=g(x). These intersection points divide the region and ensure you subtract “top minus bottom” on the right interval(s).
If f(x)≥0f(x)≥0 on [a,b][a,b], then ∫abf(x) dx∫abf(x)dx equals
A Always zero
B Always negative
C Only a slope
D Geometric area
When the curve stays above the xx-axis, the definite integral matches the geometric area under the curve between x=ax=a and x=bx=b.
If a curve crosses the xx-axis, total area is found by
A Swapping limits
B Squaring function
C Splitting at roots
D Differentiating twice
Integrals give signed area. For total area, find where f(x)=0f(x)=0, split the interval into parts where the sign is constant, and add absolute areas.
Which is the correct “linearity” property?
A ∫(af+bg)=a∫f+b∫g∫(af+bg)=a∫f+b∫g
B ∫(af)=a+∫f∫(af)=a+∫f
C ∫(fg)=∫f∫g∫(fg)=∫f∫g
D ∫(f/g)=∫f/∫g∫(f/g)=∫f/∫g
Definite integrals distribute over addition and constants can be pulled out. This is essential for breaking complicated expressions into easier integrals.
A common rationalizing substitution helps integrals containing
A sinxsinx only
B ax+bax+b
C exex only
D lnxlnx only
When radicals like ax+bax+b appear, substitutions such as t=ax+bt=ax+b can remove the square root and convert the integral into a rational form.
Which integral is typically solved using by-parts repeatedly?
A ∫x2ex dx∫x2exdx
B ∫1x dx∫x1dx
C ∫cosx dx∫cosxdx
D ∫x3 dx∫x3dx
Products of polynomials and exponentials are ideal for repeated integration by parts. Each step reduces the polynomial power until only ∫exdx∫exdx remains.
Which is correct about ∫abf(x) dx∫abf(x)dx if f(x)=0f(x)=0 always?
A Value is one
B Value is b+ab+a
C Value is zero
D Value is abab
If the function is identically zero, every rectangle height in the Riemann sum is zero. Hence the limit, the definite integral, must also be zero.
Riemann sum idea supports definite integral as
A Limit of products
B Random sampling
C Matrix inverse
D Limit of sums
A definite integral is defined as the limit of sums ∑f(xi)Δx∑f(xi)Δx as the subinterval width goes to zero. This formalizes “area under the curve.”
Which situation makes ∫abf(x) dx∫abf(x)dx negative?
A More below axis
B More above axis
C Function increasing
D Function even
Because the integral is signed area, regions where f(x)<0f(x)<0 contribute negative area. If negative magnitude dominates, the final integral becomes negative.
Which formula gives area between curves when f≥gf≥g?
A ∫ab(f+g) dx∫ab(f+g)dx
B ∫ab(fg) dx∫ab(fg)dx
C ∫ab(f−g) dx∫ab(f−g)dx
D ∫ab(f/g) dx∫ab(f/g)dx
The area between curves is “top minus bottom” integrated over correct bounds. Ensuring f(x)≥g(x)f(x)≥g(x) on the interval avoids sign confusion.
Work done by variable force F(x)F(x) uses
A F(b)−F(a)F(b)−F(a)
B ∫abF(x) dx∫abF(x)dx
C F(a)+F(b)F(a)+F(b)
D ∫abF′(x) dx∫abF′(x)dx
Small work is dW=F(x) dxdW=F(x)dx. Adding these over the path gives total work as a definite integral. This is a key physical application of integration.
Displacement from velocity v(t)v(t) on [a,b][a,b] is
A v(b)−v(a)v(b)−v(a)
B ∫abv′(t) dt∫abv′(t)dt
C ∫abv(t) dt∫abv(t)dt
D ∫abt dt∫abtdt
Displacement accumulates velocity over time. The definite integral sums small movements v(t) dtv(t)dt. If velocity changes sign, the result is net displacement.
Which is a correct standard integral?
A ∫11+x2 dx=tan−1x+C∫1+x21dx=tan−1x+C
B ∫11+x2 dx=sin−1x+C∫1+x21dx=sin−1x+C
C ∫11+x2 dx=ln∣x∣+C∫1+x21dx=ln∣x∣+C
D ∫11+x2 dx=x+C∫1+x21dx=x+C
Since ddx(tan−1x)=11+x2dxd(tan−1x)=1+x21, the antiderivative is tan−1x+Ctan−1x+C. This appears in many rational function integrals.
Which is correct for ∫11−x2 dx∫1−x21dx?
A tan−1x+Ctan−1x+C
B ln∣x∣+Cln∣x∣+C
C sec−1x+Csec−1x+C
D sin−1x+Csin−1x+C
Because ddx(sin−1x)=11−x2dxd(sin−1x)=1−x21 for ∣x∣<1∣x∣<1, the integral equals sin−1x+Csin−1x+C.
Which is a basic example of “accumulation function”?
A ∫abf(x) dx∫abf(x)dx
B f′(x)f′(x)
C ∫axf(t) dt∫axf(t)dt
D f(x)+Cf(x)+C
An accumulation function depends on upper limit xx. It measures total change from a fixed start aa up to xx, and its derivative returns f(x)f(x) under FTC.
A simple numerical integration preview method is
A Chain rule
B Product rule
C Quotient rule
D Trapezoidal rule
The trapezoidal rule estimates ∫abf(x) dx∫abf(x)dx by dividing [a,b][a,b] into subintervals and approximating each part with a trapezoid, useful when exact integration is hard.
Which statement about ∫abk dx∫abkdx is correct for constant kk?
A k(b−a)k(b−a)
B k(b+a)k(b+a)
C k2(b−a)k2(b−a)
D k/(b−a)k/(b−a)
The graph of y=ky=k forms a rectangle over [a,b][a,b]. Area equals height kk times width (b−a)(b−a), so the definite integral is k(b−a)k(b−a).
Which is the easiest way to integrate ∫2xx2+1 dx∫x2+12xdx?
A By parts method
B Partial fractions
C Substitution method
D Trig identities
Let u=x2+1u=x2+1. Then du=2x dxdu=2xdx, so the integral becomes ∫1u du=ln∣u∣+C=ln(x2+1)+C∫u1du=ln∣u∣+C=ln(x2+1)+C.
If ff is continuous, the function F(x)=∫axf(t) dtF(x)=∫axf(t)dt is
A Differentiable
B Always constant
C Always periodic
D Always linear
FTC Part I states that when ff is continuous, the accumulation function F(x)F(x) is differentiable and F′(x)=f(x)F′(x)=f(x). This links area accumulation to instantaneous rate.
Mean Value Theorem for integrals (intro) guarantees existence of cc such that
A ∫abf=f′(c)(b−a)∫abf=f′(c)(b−a)
B ∫abf=f(c)/(b−a)∫abf=f(c)/(b−a)
C ∫abf=f(a)+f(b)∫abf=f(a)+f(b)
D ∫abf=f(c)(b−a)∫abf=f(c)(b−a)
For continuous ff on [a,b][a,b], there exists cc with average value f(c)f(c). So total area equals rectangle area f(c)×(b−a)f(c)×(b−a).
A correct basic result for limits in substitution is
A Keep xx-limits
B Convert limits in uu
C Drop the bounds
D Use only midpoint
In definite integrals, after using u=g(x)u=g(x), convert x=a,bx=a,b into u=g(a),g(b)u=g(a),g(b). This avoids back-substitution and reduces mistakes in evaluation.
Which indicates the region is “bounded” for area calculation?
A Closed by curves
B Only one curve
C No intersection
D Only at infinity
A bounded region is enclosed by curves (and possibly axes/lines) forming a closed shape. Then area can be computed by setting correct limits and integrating top-minus-bottom.
Area under y=f(x)y=f(x) above x-axis from 00 to aa is
A f(a)−f(0)f(a)−f(0)
B ∫0af′(x) dx∫0af′(x)dx
C ∫0af(x) dx∫0af(x)dx
D ∫0ax dx∫0axdx
If f(x)≥0f(x)≥0 on [0,a][0,a], the definite integral directly equals the geometric area under the curve between the vertical boundaries x=0x=0 and x=ax=a.
If a function is periodic with period TT, then ∫aa+Tf(x) dx∫aa+Tf(x)dx is
A Same each period
B Always zero
C Always positive
D Always undefined
Over any full period, the integral has the same value because the function repeats. This is useful for evaluating integrals over long intervals by splitting into full periods.
Which topic is directly connected to “volume of revolution” (intro)?
A Partial derivatives
B Definite integrals
C Complex numbers
D Binomial theorem
Volume of revolution is found by adding volumes of thin slices (disks/washers/shells). This “sum of small parts” becomes a definite integral over the chosen interval.
A correct use of modulus property for area is
A ∫∣f∣=0∫∣f∣=0 always
B ∫∣f∣=∫f∫∣f∣=∫f always
C ∫∣f∣≥0∫∣f∣≥0
D ∣∫f∣=0∣∫f∣=0
Since ∣f(x)∣≥0∣f(x)∣≥0, its definite integral cannot be negative. This is important when converting signed area into total area using absolute values and splitting intervals.
An “improper definite integral” may occur when
A Integrand becomes infinite
B Function is polynomial
C Limits are equal
D Function is constant
If f(x)f(x) blows up at some point in [a,b][a,b] (like 1/(x−a)1/(x−a)), the integral is improper. It is defined using limits and may converge or diverge.
Which best describes a “reduction formula” (basic idea)?
A Converts to derivative
B Makes integral zero
C Removes constant only
D Relates higher power
A reduction formula expresses an integral with a higher power (like InIn) in terms of a lower one (like In−2In−2). This reduces repeated work for similar integrals.
Which integral most directly introduces Beta/Gamma idea (intro)?
A ∫x2 dx∫x2dx
B ∫01xm−1(1−x)n−1 dx∫01xm−1(1−x)n−1dx
C ∫ex dx∫exdx
D ∫sinx dx∫sinxdx
The Beta function is defined using a definite integral of the form ∫01xm−1(1−x)n−1dx∫01xm−1(1−x)n−1dx. It links to Gamma functions in advanced integration.
Which is a correct “polar area” preview formula?
A 12∫r2dθ21∫r2dθ
B ∫r dθ∫rdθ
C ∫r2dx∫r2dx
D 12∫r dθ21∫rdθ
In polar coordinates, small sector area is 12r2dθ21r2dθ. Summing these sectors gives total area 12∫αβr2dθ21∫αβr2dθ.
To find area of a circle by integration, a common approach is
A Use by-parts only
B Use partial fractions
C Use symmetry halves
D Use only limits
For x2+y2=r2x2+y2=r2, take y=r2−x2y=r2−x2. Compute area in one semicircle using ∫−rrr2−x2 dx∫−rrr2−x2dx and use symmetry carefully.
Integral inequalities (basic) often use which simple fact?
A If f≥gf≥g then ∫f=∫g∫f=∫g
B If f≥gf≥g then ∫f≤∫g∫f≤∫g
C If f≥gf≥g then ∫f=0∫f=0
D If f≥gf≥g then ∫f≥∫g∫f≥∫g
Definite integrals preserve order: if f(x)≥g(x)f(x)≥g(x) on [a,b][a,b], then the accumulated area under ff is at least that under gg. This supports many inequality results.