Evaluate ∫01×31+x2 dx∫011+x2x3dx using a smart algebraic split before integrating.
A 12+12ln221+21ln2
B 12−12ln221−21ln2
C ln2−12ln2−21
D 14ln241ln2
Write x31+x2=x−x1+x21+x2x3=x−1+x2x. Then integral becomes ∫01x dx−∫01×1+x2dx=12−12ln2∫01xdx−∫011+x2xdx=21−21ln2.
Compute ∫0π/2sinx1+cosx dx∫0π/21+cosxsinxdx using substitution u=1+cosxu=1+cosx.
A 1−ln21−ln2
B 12ln221ln2
C ln2ln2
D 2ln22ln2
Let u=1+cosxu=1+cosx, then du=−sinx dxdu=−sinxdx. Limits: x=0→u=2x=0→u=2, x=π/2→u=1x=π/2→u=1. Integral becomes ∫21−1udu=ln2∫21u−1du=ln2.
Evaluate ∫0π/2ln(sinx) dx∫0π/2ln(sinx)dx using symmetry with ln(cosx)ln(cosx).
A π2ln22πln2
B −π2ln2−2πln2
C −πln2−πln2
D −π4ln2−4πln2
Use I=∫0π/2ln(sinx)dx=∫0π/2ln(cosx)dxI=∫0π/2ln(sinx)dx=∫0π/2ln(cosx)dx. Add to get 2I=∫0π/2ln(12sin2x)dx=−πln2/22I=∫0π/2ln(21sin2x)dx=−πln2/2.
If F(x)=∫x2x11+t2 dtF(x)=∫x2x1+t21dt, then F′(x)F′(x) equals
A tan−1(2x)′−tan−1(x)′tan−1(2x)′−tan−1(x)′ form
B 11+4×2−11+x21+4×21−1+x21
C 21+x2−11+4×21+x22−1+4×21
D 21+4×2−11+x21+4×22−1+x21
By Leibniz rule, derivative is f(2x)⋅(2)−f(x)⋅(1)f(2x)⋅(2)−f(x)⋅(1) where f(t)=1/(1+t2)f(t)=1/(1+t2). So F′(x)=2/(1+4×2)−1/(1+x2)F′(x)=2/(1+4×2)−1/(1+x2).
Evaluate ∫01ln(1+x)x dx∫01xln(1+x)dx (standard hard result).
A π266π2
B π22424π2
C π21212π2
D ln22ln22
Use the series ln(1+x)=∑n≥1(−1)n−1xnnln(1+x)=∑n≥1(−1)n−1nxn for ∣x∣≤1∣x∣≤1. Then ∫01ln(1+x)xdx=∑n≥1(−1)n−1n2=π2/12∫01xln(1+x)dx=∑n≥1n2(−1)n−1=π2/12.
Compute ∫0111+x4 dx∫011+x41dx.
A π42+ln(1+2)2242π+22ln(1+2)
B π4242π
C ln(1+2)22ln(1+2)
D π2222π
Factor 1+x41+x4 and use partial fractions; the definite integral simplifies to a mix of arctan and log terms, giving the stated exact value.
If I=∫0π/2sinmxcosnx dxI=∫0π/2sinmxcosnxdx with mm odd, first step is
A Save one cosxcosx
B Save one sinxsinx
C Use half-angle
D Use partial fractions
When mm is odd, separate one sinxsinx factor and convert the remaining sinm−1xsinm−1x into (1−cos2x)(m−1)/2(1−cos2x)(m−1)/2. Then substitute u=cosxu=cosx.
Write sin3x=sinx(1−cos2x)sin3x=sinx(1−cos2x). Let u=cosxu=cosx, du=−sinxdxdu=−sinxdx. Limits: 0→10→1, π/2→0π/2→0. Integral becomes ∫01(u2−u4)du=1/3−1/5=2/15∫01(u2−u4)du=1/3−1/5=2/15.
Compute ∫01×21−x2 dx∫011−x2x2dx using x=sinθx=sinθ.
A π22π
B π88π
C 11
D π44π
Put x=sinθx=sinθ, dx=cosθdθdx=cosθdθ, 1−x2=cosθ1−x2=cosθ. Then integral becomes ∫0π/2sin2θ dθ=π/4∫0π/2sin2θdθ=π/4.
Evaluate ∫011−x1+x dx∫011+x1−xdx by simplifying the integrand.
A 1−2ln21−2ln2
B 2ln2−12ln2−1
C ln2ln2
D 1−ln21−ln2
1−x1+x=1−2×1+x=−1+21+x1+x1−x=1−1+x2x=−1+1+x2. Integrate: ∫01(−1)dx+2∫0111+xdx=−1+2ln2∫01(−1)dx+2∫011+x1dx=−1+2ln2. So value =2ln2−1=2ln2−1. (Thus correct option should be A; avoid sign slip.)
Let A=∫0adx1+x2A=∫0a1+x2dx. Which statement is correct?
A A=tan−1aA=tan−1a
B A=sin−1aA=sin−1a
C A=ln(1+a2)A=ln(1+a2)
D A=a1+a2A=1+a2a
The standard antiderivative is tan−1xtan−1x. So ∫0a11+x2dx=[tan−1x]0a=tan−1a∫0a1+x21dx=[tan−1x]0a=tan−1a.
Find ∫01xln(1+x) dx∫01xln(1+x)dx using integration by parts.
A 14+12ln241+21ln2
B 14−12ln241−21ln2
C 1441
D 12ln221ln2
Take u=ln(1+x)u=ln(1+x), dv=x dxdv=xdx. Then reduce to ∫x21+xdx∫1+xx2dx. Simplify x2/(1+x)=x−1+11+xx2/(1+x)=x−1+1+x1. Evaluate 0 to 1 gives 1/41/4.
Evaluate ∫01tan−1×1+x2 dx∫011+x2tan−1xdx using substitution u=tan−1xu=tan−1x.
A π21616π2
B π23232π2
C π26464π2
D π88π
Let u=tan−1xu=tan−1x, then du=dx1+x2du=1+x2dx. Limits: x=0→0x=0→0, x=1→π/4x=1→π/4. Integral becomes ∫0π/4u du=u2/2∣0π/4=π2/32∫0π/4udu=u2/2∣0π/4=π2/32.
Compute ∫01lnx1+x dx∫011+xlnxdx (classic hard integral).
A −π26−6π2
B −π224−24π2
C −ln22−ln22
D −π212−12π2
Expand 11+x=∑n≥0(−1)nxn1+x1=∑n≥0(−1)nxn on [0,1][0,1]. Then ∫01xnlnx dx=−1/(n+1)2∫01xnlnxdx=−1/(n+1)2. Summing gives −∑k≥1(−1)k−1/k2=−π2/12−∑k≥1(−1)k−1/k2=−π2/12.
Evaluate ∫0π/2dx1+sinx∫0π/21+sinxdx.
A π22π
B 22
C 11
D ln2ln2
Multiply by 1−sinx1−sinx1−sinx1−sinx: integrand becomes 1−sinxcos2x=sec2x−secxtanxcos2x1−sinx=sec2x−secxtanx. Integrate: tanx−secxtanx−secx from 0 to π/2π/2 gives 1.
Compute ∫0111−x2 dx∫011−x21dx exactly.
A ππ
B π22π
C 11
D π44π
∫11−x2dx=sin−1x+C∫1−x21dx=sin−1x+C. Evaluate from 0 to 1 gives sin−1(1)−sin−1(0)=π/2−0=π/2sin−1(1)−sin−1(0)=π/2−0=π/2.
Using a reduction idea, ∫0π/2sin4x dx∫0π/2sin4xdx equals
A 3π16163π
B π88π
C 5π32325π
D 3π883π
Use identity sin4x=(1−cos2×2)2=38−12cos2x+18cos4xsin4x=(21−cos2x)2=83−21cos2x+81cos4x. Cosine terms integrate to 0; remaining gives 38⋅π2=3π/1683⋅2π=3π/16.
Evaluate ∫01x(1+x2)2 dx∫01(1+x2)2xdx.
A 1221
B 1881
C 1441
D 3443
Let u=1+x2u=1+x2, du=2x dxdu=2xdx. Then integral becomes 12∫12u−2du=12[−u−1]12=12(1−1/2)=1/421∫12u−2du=21[−u−1]12=21(1−1/2)=1/4.
y=1−x2y=1−x2 is the upper semicircle of radius 1. From 0 to 1 it covers a quarter circle area, which equals π(1)2/4=π/4π(1)2/4=π/4.
If ff is positive decreasing, integral test compares ∑f(n)∑f(n) with
A ∫01f(x)dx∫01f(x)dx
B ∫−∞∞f(x)dx∫−∞∞f(x)dx
C ∫0∞f′(x)dx∫0∞f′(x)dx
D ∫1∞f(x)dx∫1∞f(x)dx
Integral test states: if ff is positive, continuous, decreasing on [1,∞)[1,∞), then ∑f(n)∑f(n) and ∫1∞f(x)dx∫1∞f(x)dx either both converge or both diverge.
Evaluate ∫011x dx∫01x1dx and classify it.
A 22, divergent
B 11, convergent
C 22, convergent
D 11, divergent
This is improper at x=0x=0. Compute ∫01x−1/2dx=[2×1/2]01=2∫01x−1/2dx=[2×1/2]01=2. The limit exists, so the improper integral converges.
Compute ∫01xp−1dx∫01xp−1dx converges when
A p=0p=0
B p>0p>0
C p<0p<0
D p=1p=1 only
∫01xp−1dx=[xpp]01∫01xp−1dx=[pxp]01. The term xpxp tends to 0 at 0 only when p>0p>0, giving a finite value 1/p1/p.
Evaluate ∫01dxx(1+x)∫01x(1+x)dx using x=t2x=t2.
A π/2π/2
B ππ
C ln2ln2
D 11
Let x=t2x=t2, dx=2t dtdx=2tdt, x=tx=t. Then integrand becomes ∫012tt(1+t2)dt=2∫0111+t2dt=2[tan−1t]01=π/2∫01t(1+t2)2tdt=2∫011+t21dt=2[tan−1t]01=π/2.
Compute ∫01dx1−x2∫011−x2dx as an improper integral.
A Converges to ln2ln2
B Converges to 11
C Diverges to infinity
D Converges to 00
11−x21−x21 has a vertical asymptote at x=1x=1. The integral ∫0111−x2dx∫011−x21dx involves ln∣1+x1−x∣ln1−x1+x, which blows up as x→1−x→1−.
If y=∫0x∣t−1∣dty=∫0x∣t−1∣dt, then y′(x)y′(x) for x≠1x=1 equals
A x−1x−1
B ∣x−1∣∣x−1∣
C 1−x1−x
D 00
By FTC Part I, derivative of ∫0xg(t)dt∫0xg(t)dt is g(x)g(x) where gg is continuous. Here g(t)=∣t−1∣g(t)=∣t−1∣ is continuous, so y′(x)=∣x−1∣y′(x)=∣x−1∣.
Integrate termwise: ∫01(x2−2x+1)dx=[x3/3−x2+x]01=1/3−1+1=1/3∫01(x2−2x+1)dx=[x3/3−x2+x]01=1/3−1+1=1/3. It is ∫01(x−1)2dx∫01(x−1)2dx.
Compute ∫0π/2cosx2+sinx dx∫0π/22+sinxcosxdx.
A ln23ln32
B ln2ln2
C ln32ln23
D 1−ln21−ln2
Let u=2+sinxu=2+sinx, du=cosx dxdu=cosxdx. Limits: x=0→2x=0→2, x=π/2→3x=π/2→3. Integral becomes ∫231udu=ln3−ln2=ln(3/2)∫23u1du=ln3−ln2=ln(3/2).
Evaluate ∫011(1+x)2 dx∫01(1+x)21dx.
A 1221
B ln2ln2
C 11
D 1441
Antiderivative: ∫(1+x)−2dx=−(1+x)−1∫(1+x)−2dx=−(1+x)−1. Evaluate 0 to 1: −1/2)−(−1)=1/2−1/2)−(−1)=1/2. This is a clean rational integral.
Find ∫01×21+x dx∫011+xx2dx using division.
A 12−ln221−ln2
B ln2−12ln2−21
C ln2+12ln2+21
D 1−ln21−ln2
Divide: x21+x=x−1+11+x1+xx2=x−1+1+x1. Integrate 0 to 1: ∫01(x−1)dx+∫0111+xdx=(1/2−1)+ln2=ln2−1/2∫01(x−1)dx+∫011+x1dx=(1/2−1)+ln2=ln2−1/2.
If f(x)=∫0xtet2dtf(x)=∫0xtet2dt, then f(1)f(1) equals
A e−1e−1
B 12(e+1)21(e+1)
C 12(e−1)21(e−1)
D 1−e1−e
Use substitution u=t2u=t2, du=2t dtdu=2tdt. Then f(1)=12∫01eudu=12(e−1)f(1)=21∫01eudu=21(e−1). This is a standard exponential-substitution pattern.
Evaluate ∫0π∣sinx∣dx∫0π∣sinx∣dx.
A 00
B ππ
C 11
D 22
On [0,π][0,π], sinx≥0sinx≥0, so ∣sinx∣=sinx∣sinx∣=sinx. Thus integral equals ∫0πsinxdx=2∫0πsinxdx=2. Absolute value matters on larger intervals.
Compute ∫−1111+x2dx∫−111+x21dx using symmetry.
A π22π
B ππ
C π44π
D 11
The integrand is even, so integral equals 2∫0111+x2dx=2[tan−1x]01=2(π/4)=π/22∫011+x21dx=2[tan−1x]01=2(π/4)=π/2.
Evaluate ∫0111−xdx∫011−x1dx.
A 11
B 1221
C 22
D Divergent
Improper at x=1x=1. Substitute u=1−xu=1−x, du=−dxdu=−dx. Then integral becomes ∫10u−1/2(−du)=∫01u−1/2du=[2u1/2]01=2∫10u−1/2(−du)=∫01u−1/2du=[2u1/2]01=2. Convergent.
If S=∫0π/2ln(1+sinx) dxS=∫0π/2ln(1+sinx)dx, then SS equals
A π4ln24πln2
B 2G−π2ln22G−2πln2
C π2ln22πln2
D G−π2ln2G−2πln2
Convert 1+sinx=(sinx2+cosx2)21+sinx=(sin2x+cos2x)2, reduce to ∫ln(cos)∫ln(cos) on [0,π/4][0,π/4], which involves Catalan constant GG.
Evaluate ∫0π/2ln(cosx) dx∫0π/2ln(cosx)dx.
A π2ln22πln2
B −πln2−πln2
C −π4ln2−4πln2
D −π2ln2−2πln2
It equals ∫0π/2ln(sinx)dx∫0π/2ln(sinx)dx by substitution x→π/2−xx→π/2−x. The standard evaluated value is −π2ln2−2πln2, used in many advanced integrals.
Compute ∫0111−x2dx∫011−x21dx (again) is divergent because
A Sine becomes zero
B Polynomial not integrable
C Log blows up
D Odd function cancels
Antiderivative involves 12ln∣1+x1−x∣21ln1−x1+x. As x→1−x→1−, denominator 1−x→0+1−x→0+, so logarithm goes to +∞+∞, making the improper integral diverge.
If r=2cosθr=2cosθ, area of the polar curve is
A 2π2π
B ππ
C 4π4π
D π/2π/2
Use polar area A=12∫r2dθ=12∫02π4cos2θdθ=2∫02πcos2θdθ=2⋅π=2πA=21∫r2dθ=21∫02π4cos2θdθ=2∫02πcos2θdθ=2⋅π=2π.
Compute ∫01×1−x dx∫011−xxdx as an improper integral.
A Divergent
B ln2ln2
C 11
D 00
Near x=1x=1, x1−x∼11−x1−xx∼1−x1, whose integral diverges like −ln(1−x)−ln(1−x). Thus the improper integral from 0 to 1 is divergent.
Evaluate ∫0111+x2dx∫011+x21dx exactly.
A π/2π/2
B ln2ln2
C π/4π/4
D 11
Antiderivative is tan−1xtan−1x. So value is [tan−1x]01=tan−1(1)−0=π/4[tan−1x]01=tan−1(1)−0=π/4. It is a basic but frequently used definite integral.
If M=∫011(1+x2)2dxM=∫01(1+x2)21dx, then MM equals
A 14−π841−8π
B 14+π841+8π
C 12+π821+8π
D 12−π821−8π
Use identity ∫1(1+x2)2dx=x2(1+x2)+12tan−1x+C∫(1+x2)21dx=2(1+x2)x+21tan−1x+C. Evaluate 0 to 1 gives 1/4+π/81/4+π/8.
Compute ∫0π/2dxsinx+cosx∫0π/2sinx+cosxdx.
A π22π
B π4242π
C 22
D π2222π
Write sinx+cosx=2sin(x+π/4)sinx+cosx=2sin(x+π/4). Then integral becomes 12∫0π/2csc(x+π/4)dx21∫0π/2csc(x+π/4)dx. Using symmetry gives value π/(22)π/(22).
Evaluate ∫01ln(1−x)xdx∫01xln(1−x)dx (standard hard value).
A −π212−12π2
B −π26−6π2
C π266π2
D −ln22−ln22
Use series ln(1−x)=−∑n≥1xnnln(1−x)=−∑n≥1nxn for 0
If f(x)=∫0xdt1+t4f(x)=∫0x1+t4dt, then f′(x)f′(x) equals
A 11+x41+x41
B 4×31+x41+x44x3
C tan−1(x2)tan−1(x2)
D ln(1+x4)ln(1+x4)
By FTC Part I, derivative of ∫0xg(t)dt∫0xg(t)dt is g(x)g(x), provided gg is continuous. Here g(t)=1/(1+t4)g(t)=1/(1+t4), so f′(x)=1/(1+x4)f′(x)=1/(1+x4).
Evaluate ∫01×2(1+x2)dx∫01(1+x2)x2dx quickly.
A π4−14π−1
B π44π
C 1−π41−4π
D 11
Rewrite x21+x2=1−11+x21+x2x2=1−1+x21. Integrate 0 to 1: ∫011dx−∫0111+x2dx=1−π/4∫011dx−∫011+x21dx=1−π/4.
Compute ∫01dx1+x∫011+xdx.
A 2(1−2)2(1−2)
B 2(2−1)2(2−1)
C 2−12−1
D ln2ln2
Let u=1+xu=1+x, du=dxdu=dx. Integral becomes ∫12u−1/2du=[2u1/2]12=2(2−1)∫12u−1/2du=[2u1/2]12=2(2−1). This is a clean root-substitution result.
For area between curves, why must you check intersection points?
A Integrals always zero
B Top curve fixed
C Derivative required
D Limits may change
Curves can cross, changing which one is above. Intersection points split the interval so you integrate “top minus bottom” correctly on each part, ensuring area remains positive.
Evaluate ∫0111+xdx∫011+x1dx and then square it. Which is true?
A Square not integral
B Equals π2/12π2/12
C Equals ln22ln22
D Equals 1/21/2
First integral equals ln2ln2. Squaring gives (ln2)2(ln2)2. This is not the same as integrating the square; it’s just a numerical transformation after evaluation.
If y=∫0xdt1−t2y=∫0x1−t2dt, then y(12)y(21) equals
A tan−1(1/2)tan−1(1/2)
B sin−1(1/2)sin−1(1/2)
C ln(3/2)ln(3/2)
D 1/21/2
The antiderivative is sin−1tsin−1t. So y(x)=sin−1x−sin−10=sin−1xy(x)=sin−1x−sin−10=sin−1x. Hence y(1/2)=sin−1(1/2)=π/6y(1/2)=sin−1(1/2)=π/6.