A function f(x)f(x) satisfies f′(x)=2x(1+x2)4f′(x)=2x(1+x2)4. What is ∫2x(1+x2)4 dx∫2x(1+x2)4dx?
A (1+x2)55+C5(1+x2)5+C
B (1+x2)5+C(1+x2)5+C
C 2x(1+x2)4+C2x(1+x2)4+C
D (1+x2)44+C4(1+x2)4+C
Use substitution u=1+x2u=1+x2, so du=2x dxdu=2xdx. Then integral becomes ∫u4du=u5/5+C∫u4du=u5/5+C, giving (1+x2)5/5+C(1+x2)5/5+C.
When is integration by parts most directly useful?
A Sum of powers
B Pure rational form
C Product with log
D Simple chain form
By-parts works best for products like polynomial·log, polynomial·trig, or polynomial·exponential, where choosing uu simplifies after differentiation while dvdv integrates easily.
Evaluate ∫xex dx∫xexdx (indefinite).
A xex+ex+Cxex+ex+C
B ex/x+Cex/x+C
C x2ex+Cx2ex+C
D xex−ex+Cxex−ex+C
Use by-parts: take u=xu=x, dv=exdxdv=exdx. Then du=dxdu=dx, v=exv=ex. So ∫xexdx=xex−∫exdx=xex−ex+C∫xexdx=xex−∫exdx=xex−ex+C.
For ∫3×21+x3 dx∫1+x33x2dx, best first step is
A Use partial fractions
B Substitute u=1+x3u=1+x3
C Use by parts
D Use symmetry rule
The derivative of 1+x31+x3 is 3x23x2, which is present. Substitution makes the integral ∫1u du=ln∣u∣+C∫u1du=ln∣u∣+C.
Compute ∫1×2−4 dx∫x2−41dx using partial fractions.
A 12ln∣x2−4∣+C21ln∣x2−4∣+C
B ln∣x−2∣+ln∣x+2∣+Cln∣x−2∣+ln∣x+2∣+C
C 14ln∣x−2x+2∣+C41lnx+2x−2+C
D 14ln∣x2−4∣+C41ln∣x2−4∣+C
Factor x2−4=(x−2)(x+2)x2−4=(x−2)(x+2). Decompose: 1×2−4=14(1x−2−1x+2)x2−41=41(x−21−x+21). Integrate to get the log ratio.
Evaluate ∫x1−x2 dx∫1−x2xdx.
A 1−x2+C1−x2+C
B sin−1x+Csin−1x+C
C tan−1x+Ctan−1x+C
D −1−x2+C−1−x2+C
Let u=1−x2u=1−x2, then du=−2x dxdu=−2xdx. The integral becomes −12∫u−1/2du=−u1/2+C=−1−x2+C−21∫u−1/2du=−u1/2+C=−1−x2+C.
If ff is continuous, then ddx(∫2xf(t) dt)dxd(∫2xf(t)dt) equals
A f(2)f(2)
B ∫2xf(t)dt∫2xf(t)dt
C f(x)f(x)
D f′(x)f′(x)
FTC Part I states that if F(x)=∫axf(t)dtF(x)=∫axf(t)dt with continuous ff, then F′(x)=f(x)F′(x)=f(x). The lower limit stays constant.
Evaluate ∫0πsinx dx∫0πsinxdx.
A 00
B 22
C ππ
D 11
Antiderivative of sinxsinx is −cosx−cosx. So ∫0πsinxdx=[−cosx]0π=−(−1)−(−1)=2∫0πsinxdx=[−cosx]0π=−(−1)−(−1)=2. It equals the area of one sine hump.
Using symmetry, ∫−aax2 dx∫−aax2dx equals
A 00
B ∫0ax2dx∫0ax2dx
C −2∫0ax2dx−2∫0ax2dx
D 2∫0ax2dx2∫0ax2dx
x2x2 is even, so area from [−a,0][−a,0] equals area from [0,a][0,a]. Therefore ∫−aax2dx=2∫0ax2dx∫−aax2dx=2∫0ax2dx.
Compute ∫01(2x+1) dx∫01(2x+1)dx.
A 11
B 33
C 22
D 3223
Antiderivative is x2+xx2+x. Evaluate from 0 to 1: (1+1)−0=2(1+1)−0=2. This also matches area under a line segment plus rectangle.
If F′(x)=f(x)F′(x)=f(x), then ∫abf(x) dx∫abf(x)dx equals
A F(b)−F(a)F(b)−F(a)
B F(a)−F(b)F(a)−F(b)
C F(a)+F(b)F(a)+F(b)
D F′(b)−F′(a)F′(b)−F′(a)
FTC Part II gives ∫abf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)−F(a). It works when FF is an antiderivative of ff on the interval.
For H(x)=∫1x2f(t) dtH(x)=∫1x2f(t)dt, derivative H′(x)H′(x) equals
A f(x)f(x)
B f(x2)f(x2)
C 2x f(x)2xf(x)
D 2x f(x2)2xf(x2)
Apply chain rule with FTC: derivative of ∫1u(x)f(t)dt∫1u(x)f(t)dt is f(u(x))u′(x)f(u(x))u′(x). Here u=x2u=x2, so u′=2xu′=2x.
Which integral equals ln∣x∣−1x+Cln∣x∣−x1+C?
A ∫(1x+1×2)dx∫(x1+x21)dx
B ∫(1x+x2)dx∫(x1+x2)dx
C ∫(1x−1×2)dx∫(x1−x21)dx
D ∫(1x−x2)dx∫(x1−x2)dx
∫1xdx=ln∣x∣∫x1dx=ln∣x∣. Also ∫−x−2dx=−(−x−1)=−1/x∫−x−2dx=−(−x−1)=−1/x. So the integrand must be 1x−1x2x1−x21.
Compute ∫tanxsec2x dx∫tanxsec2xdx.
A tan2×2+C2tan2x+C
B ln∣secx∣+Cln∣secx∣+C
C tanx+Ctanx+C
D secx+Csecx+C
Let u=tanxu=tanx, then du=sec2xdxdu=sec2xdx. Integral becomes ∫u du=u2/2+C=tan2x/2+C∫udu=u2/2+C=tan2x/2+C.
Evaluate ∫11+x2 dx∫1+x21dx.
A sin−1x+Csin−1x+C
B ln∣x∣+Cln∣x∣+C
C tan−1x+Ctan−1x+C
D 1x+Cx1+C
Since ddx(tan−1x)=11+x2dxd(tan−1x)=1+x21, the antiderivative is tan−1x+Ctan−1x+C. It appears in many integrals of rational forms.
For ∫02∣x−1∣ dx∫02∣x−1∣dx, correct approach is
A Use symmetry only
B Split at x=1x=1
C Use by parts
D Use partial fractions
∣x−1∣∣x−1∣ changes form at x=1x=1. Compute ∫01(1−x)dx+∫12(x−1)dx∫01(1−x)dx+∫12(x−1)dx. Splitting avoids sign mistakes and gives true area.
Area between y=xy=x and y=x2y=x2 from 00 to 11 is
A ∫01(x2−x)dx∫01(x2−x)dx
B ∫01(x+x2)dx∫01(x+x2)dx
C ∫01(x2/x)dx∫01(x2/x)dx
D ∫01(x−x2)dx∫01(x−x2)dx
On [0,1][0,1], x≥x2x≥x2. Area between curves equals ∫01[top−bottom]dx=∫01(x−x2)dx∫01[top−bottom]dx=∫01(x−x2)dx.
Compute area between y=xy=x and y=x2y=x2 on [0,1][0,1].
A 1331
B 1221
C 1661
D 2332
Area =∫01(x−x2)dx=[x2/2−x3/3]01=1/2−1/3=1/6=∫01(x−x2)dx=[x2/2−x3/3]01=1/2−1/3=1/6. This is a standard curve-area example.
For volume by disks about x-axis, the basic form is
A ∫ydx∫ydx
B π∫y2dxπ∫y2dx
C π∫ydxπ∫ydx
D ∫y2dx∫y2dx
Rotating y=f(x)≥0y=f(x)≥0 about the x-axis creates disks of radius yy. Disk volume is πy2dxπy2dx, so total volume is π∫aby2dxπ∫aby2dx.
A probability density p(x)p(x) must satisfy which basic condition?
A ∫p=0∫p=0
B p<0p<0 always
C pp periodic
D ∫p=1∫p=1
For a continuous random variable, total probability over the full range must be 1. Also p(x)≥0p(x)≥0. Probabilities on intervals are found using definite integrals.
Which integral gives arc length idea (intro) for y=f(x)y=f(x)?
A ∫1+(f′)2 dx∫1+(f′)2dx
B ∫(1+f′) dx∫(1+f′)dx
C ∫(f′)2 dx∫(f′)2dx
D ∫1+f dx∫1+fdx
Arc length sums tiny line segments. The differential length is 1+(dy/dx)2 dx1+(dy/dx)2dx. Integrating this from aa to bb gives total curve length.
If f(x)≥g(x)f(x)≥g(x) on [a,b][a,b], then area between them is
A ∫ab(g−f)dx∫ab(g−f)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
Area between curves is computed as “upper minus lower.” Keeping f≥gf≥g ensures the integrand stays nonnegative and the integral equals geometric area.
Evaluate ∫01×2 dx∫01x2dx.
A 1221
B 1441
C 2332
D 1331
Antiderivative of x2x2 is x3/3×3/3. Evaluate from 0 to 1: 1/3−0=1/31/3−0=1/3. This is a common standard definite integral value.
A correct symmetry shortcut is ∫−22(x3+1) dx∫−22(x3+1)dx equals
A ∫−22x3dx∫−22x3dx
B ∫−221dx∫−221dx
C 00
D 2∫02x3dx2∫02x3dx
Over symmetric limits, integral of odd part x3x3 is zero. Only the constant 11 remains. So ∫−22(x3+1)dx=∫−221dx=4∫−22(x3+1)dx=∫−221dx=4.
Compute ∫−22(x3+1) dx∫−22(x3+1)dx.
A 00
B 88
C 44
D −4−4
As x3x3 is odd, ∫−22x3dx=0∫−22x3dx=0. Then ∫−221dx=(2−(−2))=4∫−221dx=(2−(−2))=4. Add them to get 4.
Which integral form matches Leibniz rule (basic) idea?
A ddx∫a(x)b(x)f(t)dtdxd∫a(x)b(x)f(t)dt
B ddx∫abf(x)dxdxd∫abf(x)dx
C ∫abf′(t)dt∫abf′(t)dt
D ∫abf(x)2dx∫abf(x)2dx
Leibniz rule handles differentiation when both limits depend on xx. It combines integrand values at the moving limits with derivatives of those limits, plus any explicit xx dependence.
If J(x)=∫0xsin(t2) dtJ(x)=∫0xsin(t2)dt, then J′(x)J′(x) is
A 2xsinx2xsinx
B cos(x2)cos(x2)
C sin2xsin2x
D sin(x2)sin(x2)
By FTC Part I, derivative of ∫0xf(t)dt∫0xf(t)dt is f(x)f(x). Here f(t)=sin(t2)f(t)=sin(t2), so J′(x)=sin(x2)J′(x)=sin(x2).
Which evaluation technique is correct for ∫02(x−1)dx∫02(x−1)dx?
A Use symmetry only
B Use by-parts
C Use antiderivative
D Use partial fractions
This is a simple polynomial. Antiderivative is x2/2−xx2/2−x. Evaluate from 0 to 2: (2−2)−0=0(2−2)−0=0. Signed areas cancel around x=1x=1.
Compute ∫02(x−1) dx∫02(x−1)dx.
A 11
B 00
C 22
D −1−1
∫02(x−1)dx=[x2/2−x]02=(2−2)−0=0∫02(x−1)dx=[x2/2−x]02=(2−2)−0=0. The line is below the axis on [0,1][0,1] and above on [1,2][1,2] equally.
If f(x)≥0f(x)≥0, then which is always true?
A ∫abf≤0∫abf≤0
B ∫abf=0∫abf=0
C ∫abf=b−a∫abf=b−a
D ∫abf≥0∫abf≥0
Integrating a nonnegative function adds nonnegative contributions across the interval. So the definite integral cannot be negative, and equals zero only if the function is zero almost everywhere.
For ∫0πcosx dx∫0πcosxdx, the value is
A 22
B ππ
C 00
D −2−2
Antiderivative of cosxcosx is sinxsinx. Evaluate: sinπ−sin0=0−0=0sinπ−sin0=0−0=0. Positive and negative parts over [0,π][0,π] cancel net area.
Compute ∫1a2−x2 dx∫a2−x21dx for a>0a>0.
A sin−1(x/a)+Csin−1(x/a)+C
B tan−1(x/a)+Ctan−1(x/a)+C
C ln∣x∣+Cln∣x∣+C
D sec−1(x/a)+Csec−1(x/a)+C
Since ddxsin−1(x/a)=1a2−x2dxdsin−1(x/a)=a2−x21, the integral equals sin−1(x/a)+Csin−1(x/a)+C. This is a standard trig-substitution result.
Which is correct for ∫e2x dx∫e2xdx?
A 2e2x+C2e2x+C
B ex2+Cex2+C
C ln∣2x∣+Cln∣2x∣+C
D 12e2x+C21e2x+C
Let u=2xu=2x, so du=2dxdu=2dx. Then ∫e2xdx=12∫eudu=12eu+C=12e2x+C∫e2xdx=21∫eudu=21eu+C=21e2x+C.
Which is correct for ∫lnx dx∫lnxdx (for x>0x>0)?
A (lnx)2+C(lnx)2+C
B ln∣x∣/x+Cln∣x∣/x+C
C xlnx−x+Cxlnx−x+C
D xlnx+Cxlnx+C
Use by-parts: u=lnxu=lnx, dv=dxdv=dx. Then du=dx/xdu=dx/x, v=xv=x. So ∫lnxdx=xlnx−∫1dx=xlnx−x+C∫lnxdx=xlnx−∫1dx=xlnx−x+C.
For improper integral ∫1∞1xp dx∫1∞xp1dx, it converges when
A p>1p>1
B p=1p=1
C p<1p<1
D p=0p=0
The p-test for improper integrals says ∫1∞x−pdx∫1∞x−pdx converges if p>1p>1 and diverges if p≤1p≤1. This is a key basic convergence rule.
The integral ∫011xpdx∫01xp1dx converges when
A p=1p=1
B p>1p>1
C p<1p<1
D p=0p=0
Near x=0x=0, x−px−p is integrable only if the singularity is not too strong. The standard test gives convergence for p<1p<1 and divergence for p≥1p≥1.
A correct “change of variable” step in definite integral is
A Keep old bounds
B Ignore bounds
C Swap bounds always
D Change bounds too
In definite integrals, once you substitute u=g(x)u=g(x), you must convert x=a,bx=a,b into u=g(a),g(b)u=g(a),g(b). This keeps evaluation consistent and reduces back-substitution errors.
Which is a correct polar area preview statement?
A Uses derivative only
B Sector area formula
C Needs partial fractions
D Always zero area
In polar form, small area element resembles a sector: dA=12r2dθdA=21r2dθ. Integrating this over an angle interval gives polar region area.
Compute ∫0π/2sinx dx∫0π/2sinxdx.
A 00
B 1221
C 11
D 22
∫sinxdx=−cosx∫sinxdx=−cosx. So ∫0π/2sinxdx=[−cosx]0π/2=−(0)−(−1)=1∫0π/2sinxdx=[−cosx]0π/2=−(0)−(−1)=1. It equals area under sine in first quadrant.
For area in first quadrant bounded by y=xy=x, x-axis, and x=4x=4, area is
A ∫04x dx∫04xdx
B ∫041xdx∫04x1dx
C ∫04lnx dx∫04lnxdx
D ∫04x dx∫04xdx
The region is under the curve y=xy=x and above the x-axis from x=0x=0 to x=4x=4. Hence area equals ∫04x dx∫04xdx.
Which best describes “surface area of revolution” (intro)?
A Integrate strip areas
B Differentiate curve
C Use symmetry only
D Use factoring only
Surface area is found by summing areas of tiny frustum-like strips formed by rotation. This leads to an integral involving the curve radius and arc length element.
A correct basic idea for “moment of inertia” (intro) is
A ∫r dm∫rdm
B ∫r2dm∫r2dm
C ∫dm/r2∫dm/r2
D ∫r3dm∫r3dm
Moment of inertia measures resistance to rotation. Each mass piece contributes r2dmr2dm, where rr is distance from axis. Integrals add contributions over continuous bodies.
Which method often simplifies ∫dxx2−a2∫x2−a2dx (intro)?
A Partial fractions
B By parts
C Symmetry rule
D Trig substitution
Expressions with x2−a2x2−a2 are commonly handled using substitutions like x=asecθx=asecθ (or hyperbolic forms). This converts the root into trig expressions.
Which inequality statement is correct for integrals?
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≥∫g∫f≥∫g
D If f≥gf≥g, then ∫f=0∫f=0
Order is preserved in integration: if one function is always above another on [a,b][a,b], its accumulated area is at least as large. This helps prove many bounds.
Compute ∫1x(x+1) dx∫x(x+1)1dx.
A ln∣x∣+ln∣x+1∣+Cln∣x∣+ln∣x+1∣+C
B ln∣x∣−ln∣x+1∣+Cln∣x∣−ln∣x+1∣+C
C ln∣x+1∣+Cln∣x+1∣+C
D ln∣x∣+Cln∣x∣+C
Decompose: 1x(x+1)=1x−1x+1x(x+1)1=x1−x+11. Integrate to get ln∣x∣−ln∣x+1∣+C=ln∣xx+1∣+Cln∣x∣−ln∣x+1∣+C=lnx+1x+C.
A correct standard integral is ∫1a2+x2dx∫a2+x21dx equals
A tan−1(x/a)+Ctan−1(x/a)+C
B sin−1(x/a)+Csin−1(x/a)+C
C ln∣x+a2+x2∣+Cln∣x+a2+x2∣+C
D ln∣x∣+Cln∣x∣+C
This is a standard form derived using trig or hyperbolic substitution. Differentiating ln∣x+a2+x2∣ln∣x+a2+x2∣ gives 1/a2+x21/a2+x2, confirming correctness.
Integral test for series (intro) connects series to
A Matrix products
B Trig identities
C Vector spaces
D Improper integrals
The integral test compares ∑an∑an with ∫f(x)dx∫f(x)dx when an=f(n)an=f(n) and ff is positive, decreasing. Convergence of one implies convergence of the other.
Solve y′=2xy′=2x using integration (basic). General solution is
A y=x2+Cy=x2+C
B y=2x+Cy=2x+C
C y=x+Cy=x+C
D y=2×2+Cy=2×2+C
Integrate both sides: dy/dx=2xdy/dx=2x implies y=∫2xdx=x2+Cy=∫2xdx=x2+C. The constant CC represents the family of solution curves.
A correct “splitting interval” trick is
A ∫abf=∫acf−∫cbf∫abf=∫acf−∫cbf
B ∫abf=∫cbf∫abf=∫cbf
C ∫abf=∫acf+∫cbf∫abf=∫acf+∫cbf
D ∫abf=∫abf(c)∫abf=∫abf(c)
Definite integrals are additive across adjacent intervals. Splitting at a convenient point cc helps handle sign changes, absolute values, piecewise functions, or different simplification methods.
Which is correct for ∫01(1−x)dx∫01(1−x)dx?
A 11
B 00
C 1331
D 1221
Antiderivative is x−x2/2x−x2/2. Evaluate from 0 to 1: 1−1/2=1/21−1/2=1/2. Geometrically, it matches area of a right triangle with base 1 and height 1.