Evaluate ∫4x(1+2×2) dx∫(1+2×2)4xdx using a suitable substitution.
A ln(1+2×2)+Cln(1+2×2)+C
B ln∣x∣+Cln∣x∣+C
C 12ln(1+2×2)+C21ln(1+2×2)+C
D 14ln(1+2×2)+C41ln(1+2×2)+C
Let u=1+2x2u=1+2×2. Then du=4x dxdu=4xdx. So the integral becomes ∫1u du=ln∣u∣+C=ln(1+2×2)+C∫u1du=ln∣u∣+C=ln(1+2×2)+C. Thus coefficient is 1, but since dudu matches exactly, answer is ln(1+2×2)+Cln(1+2×2)+C. (Note: correct option should reflect this.)
Which option correctly gives ∫dxx2+4∫x2+4dx?
A tan−1(x/2)+Ctan−1(x/2)+C
B sin−1(x/2)+Csin−1(x/2)+C
C ln∣x2+4∣+Cln∣x2+4∣+C
D 12tan−1(x/2)+C21tan−1(x/2)+C
Use ∫dxx2+a2=1atan−1(x/a)+C∫x2+a2dx=a1tan−1(x/a)+C. Here a=2a=2, so result is 12tan−1(x/2)+C21tan−1(x/2)+C.
Compute ∫x2x3+1 dx∫x3+1x2dx by substitution.
A 13ln∣x3+1∣+C31ln∣x3+1∣+C
B ln∣x3+1∣+Cln∣x3+1∣+C
C 12ln∣x3+1∣+C21ln∣x3+1∣+C
D 13(x3+1)+C31(x3+1)+C
Let u=x3+1u=x3+1. Then du=3x2dxdu=3x2dx. So ∫x2x3+1dx=13∫1udu=13ln∣u∣+C∫x3+1x2dx=31∫u1du=31ln∣u∣+C.
Evaluate ∫01ln(1+x) dx∫01ln(1+x)dx using a standard approach.
A ln2−1ln2−1
B 2ln2−12ln2−1
C 1−ln21−ln2
D 2−ln22−ln2
Antiderivative: ∫ln(1+x)dx=(1+x)ln(1+x)−(1+x)+C∫ln(1+x)dx=(1+x)ln(1+x)−(1+x)+C. From 0 to 1 gives [2ln2−2]−[0−1]=2ln2−1[2ln2−2]−[0−1]=2ln2−1.
If I=∫0π/2sin2x dxI=∫0π/2sin2xdx, then II equals
A π/2π/2
B 11
C 1/21/2
D π/4π/4
Use sin2x=1−cos2x2sin2x=21−cos2x. Then I=12∫0π/21dx−12∫0π/2cos2x dx=π4I=21∫0π/21dx−21∫0π/2cos2xdx=4π.
Compute ∫1×2−1 dx∫x2−11dx.
A 12ln∣x−1x+1∣+C21lnx+1x−1+C
B ln∣x2−1∣+Cln∣x2−1∣+C
C 12ln∣x2−1∣+C21ln∣x2−1∣+C
D ln∣x+1x−1∣+Clnx−1x+1+C
Decompose: 1×2−1=12(1x−1−1x+1)x2−11=21(x−11−x+11). Integrating gives the log ratio with coefficient 1/21/2.
For F(x)=∫0sinxt2 dtF(x)=∫0sinxt2dt, F′(x)F′(x) equals
A sin2xsin2x
B 2sinxcosx2sinxcosx
C cosxcosx
D sin2xcosxsin2xcosx
First compute derivative using FTC with chain rule: derivative of ∫0u(x)t2dt∫0u(x)t2dt is u(x)2⋅u′(x)u(x)2⋅u′(x). Here u(x)=sinxu(x)=sinx, so F′(x)=sin2xcosxF′(x)=sin2xcosx.
Evaluate ∫02∣x2−1∣ dx∫02∣x2−1∣dx correctly by splitting at
A x=0x=0
B x=2x=2
C x=1x=1
D x=−1x=−1
∣x2−1∣∣x2−1∣ changes form where x2−1=0x2−1=0. On [0,2][0,2], the sign changes at x=1x=1. Split into [0,1][0,1] and [1,2][1,2] to integrate piecewise.
Area between y=2xy=2x and y=x2y=x2 on [0,2][0,2] is
A ∫02(2x−x2)dx∫02(2x−x2)dx
B ∫02(x2−2x)dx∫02(x2−2x)dx
C ∫02(2x+x2)dx∫02(2x+x2)dx
D ∫02(2x/x2)dx∫02(2x/x2)dx
Find which curve is on top: for 0≤x≤20≤x≤2, 2x≥x22x≥x2. Area equals ∫02[top−bottom]dx=∫02(2x−x2)dx∫02[top−bottom]dx=∫02(2x−x2)dx.
Compute ∫02(2x−x2) dx∫02(2x−x2)dx.
A 22
B 2332
C 44
D 4334
∫02(2x−x2)dx=[x2−x33]02=4−83=43∫02(2x−x2)dx=[x2−3×3]02=4−38=34. This is the bounded area between the line and parabola.
For y=f(x)≥0y=f(x)≥0, volume about x-axis using disks is
A ∫abf2dx∫abf2dx
B π∫abf2dxπ∫abf2dx
C π∫abfdxπ∫abfdx
D ∫abfdx∫abfdx
Rotating creates disks of radius f(x)f(x). Disk volume element is π[f(x)]2dxπ[f(x)]2dx. Summing from aa to bb gives V=π∫abf(x)2dxV=π∫abf(x)2dx.
If ff is continuous, MVT for integrals guarantees some cc with
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(c)/(b−a)∫abf=f(c)/(b−a)
D ∫abf=f(a)+f(b)∫abf=f(a)+f(b)
The theorem says a continuous function attains its average value. So there exists c∈[a,b]c∈[a,b] such that f(c)=1b−a∫abf(x)dxf(c)=b−a1∫abf(x)dx.
Evaluate ∫1(x+1)2 dx∫(x+1)21dx.
A 1x+1+Cx+11+C
B ln∣x+1∣+Cln∣x+1∣+C
C 1(x+1)3+C(x+1)31+C
D −1x+1+C−x+11+C
Write as (x+1)−2(x+1)−2. Integrate: ∫(x+1)−2dx=(x+1)−1/(−1)+C=−1x+1+C∫(x+1)−2dx=(x+1)−1/(−1)+C=−x+11+C.
Compute ∫secxtanx dx∫secxtanxdx.
A tanx+Ctanx+C
B −cscx+C−cscx+C
C secx+Csecx+C
D ln∣secx∣+Cln∣secx∣+C
Since ddx(secx)=secxtanxdxd(secx)=secxtanx, the integral is directly secx+Csecx+C. This is a standard trig integral result.
A correct substitution for ∫1+3x dx∫1+3xdx is
A u=1+3xu=1+3x
B u=xu=x
C u=1/xu=1/x
D u=lnxu=lnx
Linear inside a root suggests u=1+3xu=1+3x. Then du=3dxdu=3dx, and the integral becomes 13∫u1/2du31∫u1/2du, which is a basic power integral.
Evaluate ∫0111+x dx∫011+x1dx.
A 1−ln21−ln2
B ln2ln2
C 2ln22ln2
D 11
∫11+xdx=ln∣1+x∣+C∫1+x1dx=ln∣1+x∣+C. From 0 to 1 gives ln2−ln1=ln2ln2−ln1=ln2. It also matches area under 1/(1+x)1/(1+x).
Compute ∫01xex2 dx∫01xex2dx.
A e−1e−1
B 12(e+1)21(e+1)
C 1−e1−e
D 12(e−1)21(e−1)
Let u=x2u=x2, du=2x dxdu=2xdx. Then ∫01xex2dx=12∫01eudu=12(e1−e0)=12(e−1)∫01xex2dx=21∫01eudu=21(e1−e0)=21(e−1).
If K=∫0π/2cos2x dxK=∫0π/2cos2xdx, then KK equals
A π/4π/4
B π/2π/2
C 11
D 1/21/2
Use cos2x=1+cos2x2cos2x=21+cos2x. The cos2xcos2x term integrates to zero over [0,π/2][0,π/2]. Remaining part gives 12⋅π2=π421⋅2π=4π.
Evaluate ∫xx2+1 dx∫x2+1xdx.
A ln(x2+1)+Cln(x2+1)+C
B tan−1x+Ctan−1x+C
C −12ln(x2+1)+C−21ln(x2+1)+C
D 12ln(x2+1)+C21ln(x2+1)+C
Let u=x2+1u=x2+1, then du=2x dxdu=2xdx. So ∫xx2+1dx=12∫1udu=12ln∣u∣+C∫x2+1xdx=21∫u1du=21ln∣u∣+C.
For ∫abf(x) dx∫abf(x)dx, which statement is always true?
A Always equals f(b−a)f(b−a)
B Additivity over split
C Always positive value
D Depends only on aa
If a
Compute ∫141x dx∫14x1dx.
A 2
B 3
C 4
D 1
∫x−1/2dx=2x∫x−1/2dx=2x. So value =2(4−1)=2(2−1)=2=2(4−1)=2(2−1)=2
Which integral equals x33+C3x3+C?
A ∫x3dx∫x3dx
B ∫3x2dx∫3x2dx
C ∫1/x2dx∫1/x2dx
D ∫x2dx∫x2dx
Using power rule, ∫xndx=xn+1n+1+C∫xndx=n+1xn+1+C for n≠−1n=−1. With n=2n=2, we get x33+C3x3+C.
For ∫0πsinxcosx dx∫0πsinxcosxdx, best method is
A By parts method
B Partial fractions
C Substitution method
D Trig substitution
Let u=sinxu=sinx, then du=cosx dxdu=cosxdx. The integral becomes ∫0πu du∫0πudu. With bounds u(0)=0u(0)=0, u(π)=0u(π)=0, result becomes zero.
Compute ∫0πsinxcosx dx∫0πsinxcosxdx.
A 00
B 11
C 22
D −1−1
Using u=sinxu=sinx, the limits become 0 to 0, so the integral is 0. Also, sinxcosxsinxcosx is symmetric positive then negative over [0,π][0,π], cancelling out.
If f(x)≥g(x)f(x)≥g(x) on [a,b][a,b], then which is true?
A ∫f=∫g∫f=∫g
B ∫f≤∫g∫f≤∫g
C ∫f=0∫f=0
D ∫f≥∫g∫f≥∫g
Integration preserves order. If one function is always above another, its accumulated area is at least as large. This is a basic integral inequality used for bounds.
Compute ∫2x dx∫x2dx.
A ln∣x∣+Cln∣x∣+C
B 2ln∣x∣+C2ln∣x∣+C
C ln∣2x∣+Cln∣2x∣+C
D −2ln∣x∣+C−2ln∣x∣+C
Constant multiple rule: ∫2xdx=2∫1xdx=2ln∣x∣+C∫x2dx=2∫x1dx=2ln∣x∣+C. The absolute value handles negative xx.
Evaluate ∫02πcosx dx∫02πcosxdx.
A 22
B 11
C 00
D ππ
Antiderivative is sinxsinx. So [sinx]02π=sin(2π)−sin0=0[sinx]02π=sin(2π)−sin0=0. Over a full period, positive and negative parts cancel.
Compute ∫dxx(x+2)∫x(x+2)dx.
A ln∣xx+2∣+Clnx+2x+C
B 12ln∣x(x+2)∣+C21ln∣x(x+2)∣+C
C ln∣x+2∣+Cln∣x+2∣+C
D 12ln∣xx+2∣+C21lnx+2x+C
Decompose: 1x(x+2)=12(1x−1x+2)x(x+2)1=21(x1−x+21). Integrate to get 12(ln∣x∣−ln∣x+2∣)+C21(ln∣x∣−ln∣x+2∣)+C.
A correct integral for surface area of revolution (about x-axis) includes
A 2π∫y1+y′2dx2π∫y1+y′2dx
B π∫y2dxπ∫y2dx
C ∫1+y′dx∫1+y′dx
D 2π∫y′dx2π∫y′dx
Surface area sums small bands. Each band has circumference 2πy2πy and slant length ds=1+(dy/dx)2dxds=1+(dy/dx)2dx. Multiplying gives the standard integral.
Compute ∫01×1+x2 dx∫011+x2xdx.
A ln2ln2
B 1/21/2
C 12ln221ln2
D 1−ln21−ln2
Let u=1+x2u=1+x2, du=2xdxdu=2xdx. Then integral becomes 12∫121udu=12[lnu]12=12ln221∫12u1du=21[lnu]12=21ln2.
Which integral is best evaluated by splitting into intervals due to sign change?
A ∫x2dx∫x2dx
B ∫exdx∫exdx
C ∫cosxdx∫cosxdx
D ∫∣x−3∣dx∫∣x−3∣dx
Absolute value expressions change form at points where inside equals zero. For ∣x−3∣∣x−3∣, split at x=3x=3 to remove absolute value and integrate piecewise correctly.
Compute ∫14−x2 dx∫4−x21dx.
A sin−1(x/2)+Csin−1(x/2)+C
B tan−1(x/2)+Ctan−1(x/2)+C
C ln∣4−x2∣+Cln∣4−x2∣+C
D sec−1(x/2)+Csec−1(x/2)+C
Use the standard form ∫dxa2−x2=sin−1(x/a)+C∫a2−x2dx=sin−1(x/a)+C. Here a=2a=2, so answer is sin−1(x/2)+Csin−1(x/2)+C.
For V(x)=∫0x(1+t4) dtV(x)=∫0x(1+t4)dt, what is V(2)V(2)?
A 2+1652+516
B 2+3252+532
C 1+3251+532
D 2+852+58
V(2)=∫021dt+∫02t4dt=2+[t5/5]02=2+32/5V(2)=∫021dt+∫02t4dt=2+[t5/5]02=2+32/5. This is a direct accumulation computation.
Evaluate ∫excosx dx∫excosxdx (method choice).
A Partial fractions
B Trig substitution
C Symmetry method
D By parts twice
Products like excosxexcosx require integration by parts, often twice, to get an equation involving the original integral again. Solving that equation gives the final expression.
Which is an example of “improper due to infinity”?
A ∫1∞1x2dx∫1∞x21dx
B ∫01xdx∫01xdx
C ∫12x2dx∫12x2dx
D ∫−11xdx∫−11xdx
The upper limit is infinite, so the integral is defined using a limit. Convergence depends on whether the limit gives a finite number.
For ∫1∞1x2dx∫1∞x21dx, the value is
A 00
B 22
C 11
D ∞∞
∫1∞x−2dx=[−x−1]1∞=0−(−1)=1∫1∞x−2dx=[−x−1]1∞=0−(−1)=1. It converges because p=2>1p=2>1 in the p-test.
Which is a correct “definite integral substitution” statement?
A Substitute, keep limits
B Substitute, drop limits
C Substitute, swap always
D Substitute and change limits
With u=g(x)u=g(x), convert x=a,bx=a,b to u=g(a),g(b)u=g(a),g(b). Then integrate fully in uu. This reduces algebra mistakes and avoids back-substitution steps.
The integral ∫01xm−1(1−x)n−1dx∫01xm−1(1−x)n−1dx relates to
A Fourier series
B Beta function
C Matrix determinant
D Laplace equation
This definite integral defines the Beta function B(m,n)B(m,n) for m,n>0m,n>0. It is linked to Gamma functions by B(m,n)=Γ(m)Γ(n)/Γ(m+n)B(m,n)=Γ(m)Γ(n)/Γ(m+n).
Polar region area from θ=αθ=α to ββ is given by
A 12∫αβr2dθ21∫αβr2dθ
B ∫αβrdθ∫αβrdθ
C ∫αβr2dx∫αβr2dx
D 12∫αβrdθ21∫αβrdθ
A small sector has area 12r2dθ21r2dθ. Adding sectors over an interval gives total area 12∫αβr2dθ21∫αβr2dθ.
A correct idea of numerical integration is
A Differentiate repeatedly
B Always exact answer
C Use only symmetry
D Approximate with trapezoids
Numerical methods estimate ∫abf(x)dx∫abf(x)dx using simple shapes. The trapezoidal rule uses trapezoids on subintervals, improving accuracy with more partitions.
If f(x)=g(x)f(x)=g(x) on [a,b][a,b], then ∫ab(f−g)dx∫ab(f−g)dx equals
A 11
B b−ab−a
C 00
D f(b)−f(a)f(b)−f(a)
If two functions are equal everywhere on the interval, their difference is zero everywhere. The integral of the zero function over any interval is zero.
Compute ∫01(3×2) dx∫01(3×2)dx.
A 33
B 1/31/3
C 22
D 11
Antiderivative of 3x23x2 is x3x3. Evaluate from 0 to 1 gives 1−0=11−0=1. This is a basic example of reversing differentiation.
Which is correct for ∫1xx2−1dx∫xx2−11dx (intro)?
A sin−1x+Csin−1x+C
B sec−1∣x∣+Csec−1∣x∣+C
C tan−1x+Ctan−1x+C
D ln∣x∣+Cln∣x∣+C
A standard result is ddx(sec−1∣x∣)=1∣x∣x2−1dxd(sec−1∣x∣)=∣x∣x2−11. For x>1x>1, this reduces to 1/(xx2−1)1/(xx2−1), matching the integrand.
Which is a correct integral inequality fact?
A If f≥0f≥0, then ∫f≥0∫f≥0
B If f≥0f≥0, then ∫f≤0∫f≤0
C If f≥0f≥0, then ∫f=0∫f=0
D If f≥0f≥0, then ∫f=b−a∫f=b−a
Nonnegative functions contribute nonnegative area everywhere. Therefore the definite integral over any interval cannot be negative. It becomes zero only if the function is zero almost everywhere.
A correct definite integral property for constant kk is
A ∫abkf=∫abf+k∫abkf=∫abf+k
B ∫abkf=k+∫abf∫abkf=k+∫abf
C ∫abkf=k∫abf∫abkf=k∫abf
D ∫abkf=(∫abf)k∫abkf=(∫abf)k
Constants factor out of integrals because Riemann sums scale by the constant. This linearity property is basic and is used in simplifying many definite integral problems.
When does ∫abf(x)dx∫abf(x)dx equal geometric area?
A f(x)≥0f(x)≥0
B ff is odd
C Limits symmetric
D ff is decreasing
If the curve never goes below the x-axis on [a,b][a,b], the signed area equals the actual area. If it crosses, you must split intervals and use absolute values.
Evaluate ∫0π/2cosx dx∫0π/2cosxdx.
A 00
B 22
C π/2π/2
D 11
Antiderivative of cosxcosx is sinxsinx. Evaluate from 0 to π/2π/2: sin(π/2)−sin0=1−0=1sin(π/2)−sin0=1−0=1.
Which is the correct idea of “definite integral as limit”?
A Limit of derivatives
B Limit of Riemann sums
C Limit of products
D Limit of matrices
Definite integral is defined as limn→∞∑f(xi)Δxlimn→∞∑f(xi)Δx. This formal definition explains why integrals represent accumulated area or total change.
If P(x)=∫0x(t2+2t)dtP(x)=∫0x(t2+2t)dt, then P(1)P(1) equals
A 22
B 5335
C 11
D 4334
P(1)=∫01(t2+2t)dt=[t3/3+t2]01=1/3+1=4/3P(1)=∫01(t2+2t)dt=[t3/3+t2]01=1/3+1=4/3. This uses basic definite integration and accumulation.