Which statement best defines an antiderivative of f(x)f(x)?
A Function is constant
B Integral equals zero
C Derivative equals f(x)f(x)
D Graph is symmetric
An antiderivative F(x)F(x) satisfies F′(x)=f(x)F′(x)=f(x). Indefinite integration finds a family of such functions, differing only by an arbitrary constant CC.
What does the constant CC represent in ∫f(x) dx=F(x)+C∫f(x)dx=F(x)+C?
A One solution family
B Area under curve
C Limit of sum
D A slope value
Different antiderivatives of the same function differ by a constant. The CC collects all possible vertical shifts, giving the full family of solutions.
Which is the basic rule for ∫xn dx∫xndx for n≠−1n=−1?
A nxn−1+Cnxn−1+C
B xn+1n+1+Cn+1xn+1+C
C lnx+Clnx+C
D 1xn+Cxn1+C
The power rule reverses differentiation: increase the power by 1 and divide by the new power, valid for all real nn except −1−1.
Which integral equals ln∣x∣+Cln∣x∣+C?
A ∫x dx∫xdx
B ∫1x dx∫x1dx
C ∫1x dx∫x1dx
D ∫sinx dx∫sinxdx
Since ddx(ln∣x∣)=1xdxd(ln∣x∣)=x1 for x≠0x=0, the antiderivative of 1/x1/x is ln∣x∣+Cln∣x∣+C.
Which method is best for ∫2xcos(x2) dx∫2xcos(x2)dx?
A Substitution method
B Partial fractions
C By parts
D Trig identities
Let u=x2u=x2, then du=2x dxdu=2xdx. The integral becomes ∫cosu du=sinu+C=sin(x2)+C∫cosudu=sinu+C=sin(x2)+C.
In substitution, what should uu usually be?
A A derivative only
B Any constant
C Final answer
D Inner expression
Choose uu as an inner function whose derivative appears (up to a constant) in the integrand. This converts the integral into a simpler basic form.
Which method suits ∫xex dx∫xexdx?
A Integration by parts
B Partial fractions
C Trig substitution
D Symmetry property
It’s a product of algebraic and exponential terms. Using by-parts with u=xu=x, dv=exdxdv=exdx reduces it to xex−∫exdxxex−∫exdx.
What is the by-parts formula?
A ∫u dv=vu∫udv=vu
B ∫u dv=uv+∫v du∫udv=uv+∫vdu
C ∫u dv=uv−∫v du∫udv=uv−∫vdu
D ∫u dv=u+v∫udv=u+v
Integration by parts comes from the product rule: (uv)′=u′v+uv′(uv)′=u′v+uv′. Rearranging and integrating gives the standard formula used to simplify products.
Which technique is needed for ∫1×2−1 dx∫x2−11dx?
A Partial fractions
B Partial fractions
C By parts
D Area formula
Rational functions with factorable denominators are split into simpler fractions. Here x2−1=(x−1)(x+1)x2−1=(x−1)(x+1), so decomposition leads to logarithmic terms.
Partial fractions mainly applies to which integrals?
A Rational functions
B Only trig forms
C Only exponentials
D Only definite areas
Partial fractions breaks a rational function P(x)/Q(x)P(x)/Q(x) into simpler pieces whose integrals are standard, usually after ensuring the fraction is proper.
Which identity helps integrate sin2xsin2x?
A Pythagoras only
B Half-angle identity
C Half-angle identity
D Chain rule
Use sin2x=1−cos2x2sin2x=21−cos2x. This converts a squared trig function into simpler terms that integrate directly.
Which is a correct standard integral?
A ∫tanx+C∫exdx=tanx+C
B ∫exdx=lnx+C∫exdx=lnx+C
C ∫exdx=x2+C∫exdx=x2+C
D ∫exdx=ex+C∫exdx=ex+C
Exponential exex is its own derivative, so its antiderivative is also exex. This is one of the most used standard results.
Which integral gives −cosx+C−cosx+C?
A ∫cosx dx∫cosxdx
B ∫sinx dx∫sinxdx
C ∫sec2x dx∫sec2xdx
D ∫csc2x dx∫csc2xdx
Since ddx(cosx)=−sinxdxd(cosx)=−sinx, the antiderivative of sinxsinx is −cosx+C−cosx+C.
Which integral equals tanx+Ctanx+C?
A ∫sinx dx∫sinxdx
B ∫sec2x dx∫sec2xdx
C ∫sec2x dx∫sec2xdx
D ∫secx dx∫secxdx
ddx(tanx)=sec2xdxd(tanx)=sec2x. This direct derivative–antiderivative pair is frequently used in substitution problems too.
What does a definite integral ∫abf(x) dx∫abf(x)dx represent?
A Signed area
B Only slope
C Only maximum
D Always positive
A definite integral measures net accumulation: area above the xx-axis counts positive and below counts negative. For pure geometric area, absolute value or splitting is needed.
Which property is always true?
A ∫abf(x) dx=b−a∫abf(x)dx=b−a
B ∫abf(x) dx=f(a)∫abf(x)dx=f(a)
C ∫aaf(x) dx=0∫aaf(x)dx=0
D ∫abf(x) dx=0∫abf(x)dx=0
Over a zero-length interval, no accumulation occurs. So the definite integral from a point to itself is always zero for any integrable function.
What happens when limits are swapped?
A Sign changes
B Value squares
C Value doubles
D No change
∫abf(x) dx=−∫baf(x) dx∫abf(x)dx=−∫baf(x)dx. Reversing direction flips the sign because accumulation is measured with orientation.
FTC Part I mainly links integrals with what?
A Derivatives
B Derivatives
C Sequences
D Vectors
FTC Part I says if F(x)=∫axf(t) dtF(x)=∫axf(t)dt with suitable conditions, then F′(x)=f(x)F′(x)=f(x). It connects area accumulation to instantaneous rate.
FTC Part II helps evaluate ∫abf(x) dx∫abf(x)dx using what?
A Graph sketch only
B Antiderivatives
C Antiderivatives
D Factorization only
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 converts area/accumulation into simple substitution of bounds.
For F(x)=∫0x(t2+1) dtF(x)=∫0x(t2+1)dt, what is F′(x)F′(x)?
A x2+1×2+1
B 2x+12x+1
C x3x3
D t2+1t2+1
By FTC Part I, derivative of an integral with upper limit xx equals the integrand evaluated at xx. So F′(x)=x2+1F′(x)=x2+1.
Which is a key condition for FTC (basic)?
A ff must be odd
B ff must be linear
C Interval must be infinite
D Continuity of ff
In basic calculus, FTC is stated for continuous functions on [a,b][a,b]. Continuity ensures integrability and smooth behavior of the accumulation function.
If ff is even, then ∫−aaf(x) dx∫−aaf(x)dx equals
A 00
B 2∫0af(x) dx2∫0af(x)dx
C 2∫0af(x) dx2∫0af(x)dx
D −2∫0af(x) dx−2∫0af(x)dx
Even functions satisfy f(−x)=f(x)f(−x)=f(x). Areas from [−a,0][−a,0] and [0,a][0,a] match, so the total integral doubles the half-interval integral.
If ff is odd, then ∫−aaf(x) dx∫−aaf(x)dx equals
A 00
B 2∫0af(x) dx2∫0af(x)dx
C ∫0af(x) dx∫0af(x)dx
D Always positive
Odd functions satisfy f(−x)=−f(x)f(−x)=−f(x). Positive and negative contributions cancel symmetrically over [−a,a][−a,a], giving net integral zero.
Which substitution fits ∫1−x2 dx∫1−x2dx?
A x=eθx=eθ
B x=sinθx=sinθ
C x=θ2x=θ2
D x=lnθx=lnθ
For 1−x21−x2, using x=sinθx=sinθ makes 1−x2=cos2θ1−x2=cos2θ. The root simplifies and the integral becomes a trig integral.
Which integral is improper?
A ∫01x dx∫01xdx
B ∫1∞1×2 dx∫1∞x21dx
C ∫1∞1×2 dx∫1∞x21dx
D ∫−11x dx∫−11xdx
An improper integral has infinite limits or an infinite discontinuity. Here the upper limit is infinity, so it must be evaluated using a limit process.
What does “improper integral” require (intro)?
A Limits of integrals
B Only graphing
C Only factoring
D Only symmetry
Improper integrals are defined by limits because the interval is unbounded or the integrand blows up. Convergence depends on whether the limit gives a finite value.
Area between curves y=f(x)y=f(x) and y=g(x)y=g(x) on [a,b][a,b] is
A ∫ab(f+g) dx∫ab(f+g)dx
B ∫ab(f−g) dx∫ab(f−g)dx
C ∫ab(fg) dx∫ab(fg)dx
D ∫ab(f/g) dx∫ab(f/g)dx
When f(x)≥g(x)f(x)≥g(x) on [a,b][a,b], the area between them is ∫ab[f(x)−g(x)]dx∫ab[f(x)−g(x)]dx. Intersection points set correct bounds.
First step to find area between two curves is
A Differentiate both
B Find intersection points
C Find intersection points
D Use partial fractions
Area limits come from where curves meet. Solving f(x)=g(x)f(x)=g(x) gives intersection points, which define the interval(s) over which one curve stays above the other.
Which integral gives geometric area under y=f(x)≥0y=f(x)≥0 from aa to bb?
A ∫abf(x) dx∫abf(x)dx
B ∫abf′(x) dx∫abf′(x)dx
C ∫ab∣f′(x)∣ dx∫ab∣f′(x)∣dx
D ∫abf(x)2 dx∫abf(x)2dx
If the function stays above the xx-axis, the definite integral directly equals the geometric area under the curve on that interval. No absolute value is needed.
For area with x-axis when f(x)f(x) changes sign, you should
A Differentiate first
B Swap the limits
C Square the function
D Split at zeros
If f(x)f(x) crosses the axis, the integral gives signed area. To get total geometric area, split at roots and sum absolute contributions on each subinterval.
Average value of f(x)f(x) on [a,b][a,b] is
A ∫abf(x) dx∫abf(x)dx
B f(a)+f(b)22f(a)+f(b)
C 1b−a∫abf(x) dxb−a1∫abf(x)dx
D ∫abf′(x) dx∫abf′(x)dx
The average value formula divides total accumulation by interval length. It matches the height of a rectangle with the same base (b−a)(b−a) and equal area.
Work done by variable force F(x)F(x) from aa to bb is
A ∫abx dx∫abxdx
B ∫abF(x) dx∫abF(x)dx
C ∫abF′(x) dx∫abF′(x)dx
D F(b)−F(a)F(b)−F(a)
When force varies with position, small work is dW=F(x) dxdW=F(x)dx. Summing over the interval gives total work as the definite integral.
If velocity is v(t)v(t), displacement from t=at=a to t=bt=b is
A ∫abv(t) dt∫abv(t)dt
B ∫abv′(t) dt∫abv′(t)dt
C v(b)−v(a)v(b)−v(a)
D ∫abt dt∫abtdt
Displacement accumulates velocity over time. The integral adds infinitesimal movements v(t) dtv(t)dt. If direction changes, the result is signed displacement.
Which expression represents “accumulation function” example?
A f(x)2f(x)2
B ∫abf(x) dx∫abf(x)dx
C f′(x)f′(x)
D ∫axf(t) dt∫axf(t)dt
An accumulation function depends on the upper limit xx. It measures total accumulated quantity from a fixed start aa up to the variable endpoint xx.
Basic property of definite integrals is linearity
A ∫(af+bg)=ab∫f∫g∫(af+bg)=ab∫f∫g
B ∫(af+bg)=a∫f+b∫g∫(af+bg)=a∫f+b∫g
C ∫(af+bg)=a∫f+b∫g∫(af+bg)=a∫f+b∫g
D ∫(af+bg)=0∫(af+bg)=0
Definite integrals distribute over addition and constants pull out. This makes many problems easier by splitting complicated integrals into simpler ones.
Which is true for any constant kk?
A ∫abk dx=k(b−a)∫abkdx=k(b−a)
B ∫abk dx=k(b+a)∫abkdx=k(b+a)
C ∫abk dx=0∫abkdx=0
D ∫abk dx=k2∫abkdx=k2
The area under a constant function is a rectangle: height kk and width (b−a)(b−a). So the integral equals k⋅(b−a)k⋅(b−a).
Which statement about substitution in definite integrals is correct?
A Keep same limits
B Change limits to uu-values
C Replace with derivatives
D Ignore the bounds
When using u=g(x)u=g(x) in a definite integral, you must convert x=a,bx=a,b into corresponding uu-limits. Then evaluate fully in uu without back-substitution.
A quick symmetry result is ∫02πsinx dx∫02πsinxdx equals
A 11
B 00
C 00
D −1−1
Over a full period [0,2π][0,2π], sinxsinx has equal positive and negative areas. The net signed area cancels, giving zero.
Which is a standard definite integral value?
A ∫01x dx=12∫01xdx=21
B ∫01x dx=1∫01xdx=1
C ∫01x dx=0∫01xdx=0
D ∫01x dx=2∫01xdx=2
Antiderivative of xx is x2/2×2/2. Evaluate from 0 to 1: 12/2−0=1/212/2−0=1/2. It also matches triangle area under y=xy=x.
What is ∫0a1 dx∫0a1dx?
A aa
B 11
C 00
D a2a2
Integrating 1 gives length of the interval. Geometrically it’s a rectangle of height 1 and width aa, so the area (integral) equals aa.
When does ∫abf(x) dx∫abf(x)dx equal negative?
A Curve is increasing
B More area below axis
C Function is even
D Limits are equal
The definite integral is net signed area. If the region where f(x)f(x) is negative contributes greater magnitude than the positive part, the total becomes negative.
Which term describes ∫abf(x) dx∫abf(x)dx as a sum limit?
A Harmonic mean
B Binomial sum
C Riemann sum
D Vector product
A definite integral can be defined as the limit of Riemann sums: adding f(xi)Δxf(xi)Δx over many small subintervals, as Δx→0Δx→0.
Which is the correct idea of “definite integral as area”?
A Area under curve
B Slope at point
C Tangent equation
D Area under curve
For nonnegative f(x)f(x), ∫abf(x) dx∫abf(x)dx equals the geometric area between the curve, the xx-axis, and the vertical lines x=ax=a and x=bx=b.
Arc length and surface area are examples of
A Applications of integrals
B Partial fractions only
C FTC conditions
D Trig identities only
Integrals model accumulation of small pieces. Arc length sums tiny line segments, and surface area sums small surface patches, both derived using limits that become definite integrals.
Volume of revolution is usually found using
A Only derivatives
B Definite integrals
C Definite integrals
D Only matrices
Solids of revolution are built from thin disks, washers, or shells. Adding volumes of these slices over an interval leads to a definite integral formula.
Which is a correct reduction-formula idea (basic)?
A Replaces integral by derivative
B Relates InIn to In−2In−2
C Makes integral always zero
D Works only for constants
Reduction formulas connect integrals of higher powers to lower powers, such as ∫sinnx dx∫sinnxdx expressed using ∫sinn−2x dx∫sinn−2xdx, making repeated evaluation easier.
Which function’s integral often uses substitution u=lnxu=lnx?
A ∫1x dx∫x1dx style
B ∫x2 dx∫x2dx style
C ∫sinx dx∫sinxdx style
D ∫1 dx∫1dx style
Whenever the integrand contains 1xx1 along with a function of lnxlnx, the substitution u=lnxu=lnx works well because du=1xdxdu=x1dx.
A common trick for ∫1a2−x2 dx∫a2−x21dx is
A Partial fractions
B Trig substitution
C Trig substitution
D Long division
For a2−x2a2−x2, set x=asinθx=asinθ. Then the root becomes acosθacosθ, simplifying the integrand into a basic trig form.
Which statement about ∫∣f(x)∣dx∫∣f(x)∣dx is correct (basic)?
A Split where f=0f=0
B Always equals zero
C Always equals ∫f∫f
D Ignore absolute value
Absolute value changes sign handling. To integrate ∣f(x)∣∣f(x)∣, find where f(x)f(x) is positive or negative, split intervals at zeros, then remove absolute value piecewise.
Which is an example of numerical integration method (preview)?
A Quotient rule
B Product rule
C Chain rule
D Trapezoidal rule
Numerical integration estimates a definite integral using simple shapes. The trapezoidal rule approximates the area by trapezoids over subintervals, useful when exact antiderivatives are hard.