f'(t)=41+t2,f(1)=0f′(t)=41+t2,f(1)=0 Find ff.

# f'(t)=41+t2,f(1)=0f′(t)=41+t2,f(1)=0 Find ff.

• MATH

g'(x)=4×2,g(−1)=3g′(x)=4×2,g(-1)=3 Find the particular solution that satisfies the differential…

You need to use direct integration to evaluate the general solution to the differential equation: ∫(4×2)dx=4×33+c∫(4×2)dx=4×33+c You need to find the particular solution using the information…

• MATH

f'(x)−6x,f(0)=8f′(x)-6x,f(0)=8 Find the particular solution that satisfies the differential equation.

To find the solution of the differential equation f'(x)=6xf′(x)=6x  with initial condition f(0)=8f(0)=8 , integrate both sides: ∫(f'(x)dx)=∫(6xdx)⇒f(x)+c1=3×2+c2∫(f′(x)dx)=∫(6xdx)⇒f(x)+c1=3×2+c2 ,where c1,c2c1,c2are…

• MATH

∫(4x−csc2(x))dx∫(4x-csc2(x))dx Find the indefinite integral.

You need to evaluate the indefinite integral, such that: ∫(4x−csc2x)dx=∫4xdx−∫csc2xdx∫(4x-csc2x)dx=∫4xdx-∫csc2xdx ∫(4x−csc2x)dx=∫4xdx−∫1sin2xdx∫(4x-csc2x)dx=∫4xdx-∫1sin2xdx You need to remember that

• MATH

∫(tan2(y)+1)dy∫(tan2(y)+1)dy Find the indefinite integral.

You need to evaluate the indefinite integral and yo need to use the following trigonometric formula 1+tan2y=1cos2y.1+tan2y=1cos2y. Replacing 1cos2y1cos2y for 1+tan2y1+tan2y yields:

• MATH

∫sec(y)(tan(y)−sec(y))dy∫sec(y)(tan(y)-sec(y))dy Find the indefinite integral.

You need to evaluate the indefinite integral, such that: ∫secy(tany−secy)dy=∫siny−1cos2ydy=∫sinycos2ydy−∫1cos2ydy∫secy(tany-secy)dy=∫siny-1cos2ydy=∫sinycos2ydy-∫1cos2ydyYou need to solve

• MATH

∫(sec2(θ)−sin(θ))dthη∫(sec2(θ)-sin(θ))dthη Find the indefinite integral.

You need to evaluate the indefinite integral, such that: ∫(sec2θ−sinθ)dθ=∫sec2θdθ−∫sinθdθ∫(sec2θ-sinθ)dθ=∫sec2θdθ-∫sinθdθ

• MATH

∫(θ2+sec2(θ))dthη∫(θ2+sec2(θ))dthη Find the indefinite integral.

It is ∫(t2)dt+∫(sec2(t))dt=t33+tan(t)+C.∫(t2)dt+∫(sec2(t))dt=t33+tan(t)+C. where C is any constant. Both integrals are from must know table.

• MATH

∫(1−csc(t)cot(t))dt∫(1-csc(t)cot(t))dt Find the indefinite integral.

∫(1−csc(c)cot(t))dt∫(1-csc(c)cot(t))dt apply the sum rule, =∫1dt−∫csc(t)cot(t)dt=∫1dt-∫csc(t)cot(t)dt =t−∫((1sin(t)(cos(t)sin(t)))dt=t-∫((1sin(t)(cos(t)sin(t)))dt =t−∫cos(t)sin2(t)dt=t-∫cos(t)sin2(t)dt Now, let sin(t)=x ⇒⇒ cos(t)dt=dx =t−∫dxx2=t-∫dxx2apply…

• MATH

∫(t2−cos(t))dt∫(t2-cos(t))dt Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to split the integral, such that: ∫(t2−cost)dt=∫t2dt−∫costdt∫(t2-cost)dt=∫t2dt-∫costdt You need to use the following formula

• MATH

∫(5cos(x)+4sin(x))dx∫(5cos(x)+4sin(x))dx Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to split the integral, such that: ∫(5cosx+4sinx)dx=∫5cosx⋅dx+∫4sinx⋅dx∫(5cosx+4sinx)dx=∫5cosx⋅dx+∫4sinx⋅dx ∫5cosx⋅dx=5sinx+c∫5cosx⋅dx=5sinx+c

• MATH

∫(4t2+3)2dx∫(4t2+3)2dx Find the indefinite integral.

∫(4t2+3)2dt∫(4t2+3)2dt =∫(16t4+24t2+9)dt=∫(16t4+24t2+9)dt apply the sum rule and power rule, =∫16t4dt+∫24t2dt+∫9dt=∫16t4dt+∫24t2dt+∫9dt =16(t4+14+1)+24(t2+12+1)+9t=16(t4+14+1)+24(t2+12+1)+9t =16t55+8t3+9t+C=16t55+8t3+9t+C C is constant

• MATH

∫(x+1)(3x−2)dx∫(x+1)(3x-2)dx Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to open the brackets such that: (x+1)(3x−2)=3×2−2x+3x−2=3×2+x−2(x+1)(3x-2)=3×2-2x+3x-2=3×2+x-2 ∫(x+1)(3x−2)dx=∫(3×2+x−2)dx∫(x+1)(3x-2)dx=∫(3×2+x-2)dx You…

• MATH

∫(x4−3×2+5×4)dx∫(x4-3×2+5×4)dx Find the indefinite integral.

∫(x4−3×2+5×4)dx∫(x4-3×2+5×4)dx apply the sum rule =∫x4x4dx−∫3x2x4dx+∫5x4dx=∫x4x4dx-∫3x2x4dx+∫5x4dx =∫1dx−∫3x−2dx+∫5x−4dx=∫1dx-∫3x-2dx+∫5x-4dx =x−3(x−2+1−2+1)+5(x−4+1−4+1)=x-3(x-2+1-2+1)+5(x-4+1-4+1) =x+3x−1+5x−3−3=x+3x-1+5x-3-3…

• MATH

∫(x+6x−−√)dx∫(x+6x)dx Find the indefinite integral.

∫(x+6x−−√)dx∫(x+6x)dx apply the sum rule =∫(xx−−√)dx+∫(6x−−√)dx=∫(xx)dx+∫(6x)dx Now, ∫(xx−−√)dx=∫x−−√dx∫(xx)dx=∫xdxapply the power rule =x12+112+1=x12+112+1 =23×32=23×32…

• MATH

∫(3×7)dx∫(3×7)dx Find the indefinite integral.

You need to find the indefinite integral, hence you need to use the following formula: ∫1xndx=x−n+11−n∫1xndx=x-n+11-n Replacing 7 for n yields: ∫3x7dx=3x−7+11−7∫3x7dx=3x-7+11-7

• MATH

∫(1×5)dx∫(1×5)dx Find the indefinite integral.

You need to find the indefinite integral, hence you need to use the following formula: ∫1xndx=x−n+11−n∫1xndx=x-n+11-n Replacing 5 for n yields: ∫1x5dx=x−5+11−5∫1x5dx=x-5+11-5

• MATH

∫(x3−−√4+1)dx∫(x34+1)dx Find the indefinite integral.

You need to evaluate the indefinite integral, such that: ∫(x3−−√4+1)dx=∫x3−−√4dx+∫dx∫(x34+1)dx=∫x34dx+∫dx You may use the following formula ∫xndx=xn+1n+1+c∫xndx=xn+1n+1+c

• MATH

∫(x2−−√3)dx∫(x23)dx Find the indefinite integral.

You need to find the indefinite integral, hence you need to use the following formula: ∫xndx=xn+1n+1+c∫xndx=xn+1n+1+c Replacing 2323 for n yields: ∫x23dx=x23+123+1+c∫x23dx=x23+123+1+c…

• MATH

∫(x−−√+12x−−√)dx∫(x+12x)dx Find the indefinite integral.

∫(x−−√+12x−−√)dx∫(x+12x)dx apply the sum rule =∫x−−√dx+∫12x−−√dx=∫xdx+∫12xdxNow, ∫x−−√dx∫xdx apply the power rule =x12+112+1=x12+112+1 =23×32=23×32 ∫12x−−√dx∫12xdx…

• MATH

∫(x32+2x+1)dx∫(x32+2x+1)dx Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to split the integral, such that: ∫(x32+2x+1)dx=∫x32dx+∫2xdx+∫dx∫(x32+2x+1)dx=∫x32dx+∫2xdx+∫dx You need to use the following…

• MATH

∫(8×3−9×2+4)dx∫(8×3-9×2+4)dx Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to split the integral, such that: ∫(8×3−9×2+4)dx=∫8x3dx−∫9x2dx+∫4dx∫(8×3-9×2+4)dx=∫8x3dx-∫9x2dx+∫4dx You need to use the following…

• MATH

∫(x5+1)dx∫(x5+1)dx Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to split the integral, such that: ∫(x5+1)dx=∫x5dx+∫dx∫(x5+1)dx=∫x5dx+∫dx You need to use the following formula

• MATH

∫(13−x)dx∫(13-x)dx Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to split the integral, such that: ∫(13−x)dx=∫13dx−∫xdx∫(13-x)dx=∫13dx-∫xdx You need to use the following formula

• MATH

∫(x+7)dx∫(x+7)dx Find the indefinite integral.

You need to evaluate the indefinite integral, hence, you need to split the integral, such that: ∫(x+7)dx=∫xdx−∫7dx∫(x+7)dx=∫xdx-∫7dx You need to use the following formula

• SOCIAL SCIENCES

Question 1: The best answer is D. All of these groups or individuals can have some influence over the economy of the United States. The greatest influence on the economy is held by the people….

• MATH

a(t)=t2−4t+6,s(0)=0,s(1)=20a(t)=t2-4t+6,s(0)=0,s(1)=20 A particle is moving with the given data. Find…

a(t)=t2−4t+6a(t)=t2-4t+6 a(t)=v'(t)a(t)=v′(t) v(t)=∫a(t)dtv(t)=∫a(t)dt v(t)=∫(t2−4t+6)dtv(t)=∫(t2-4t+6)dt v(t)=t33−4(t22)+6t+c1v(t)=t33-4(t22)+6t+c1 v(t)=t33−2t2+6t+c1v(t)=t33-2t2+6t+c1 s(t)=∫v(t)dts(t)=∫v(t)dt s(t)=∫(t33)−2t2+6t+c1)dts(t)=∫(t33)-2t2+6t+c1)dt…

• MATH

a(t)=10sin(t)+3cos(t),s(0)=0,s(2π)=12a(t)=10sin(t)+3cos(t),s(0)=0,s(2π)=12 A particle is moving with the given data….

a(t)=10sin(t)+3cos(t)a(t)=10sin(t)+3cos(t) a(t)=v'(t)a(t)=v′(t) v(t)=∫a(t)dtv(t)=∫a(t)dt v(t)=∫(10sin(t)+3cos(t))dtv(t)=∫(10sin(t)+3cos(t))dt v(t)=−10cos(t)+3sin(t)+c1v(t)=-10cos(t)+3sin(t)+c1 v(t)=s'(t)v(t)=s′(t) s(t)=∫v(t)dts(t)=∫v(t)dt s(t)=∫(−10cos(t)+3sin(t)+c1)dts(t)=∫(-10cos(t)+3sin(t)+c1)dt…

• MATH

a(t)=3cos(t)−2sin(t),s(0)=0,v(0)=4a(t)=3cos(t)-2sin(t),s(0)=0,v(0)=4 A particle is moving with the given data….

You need to remember the relation between acceleration, velocity and position, such that: ∫a(t)dt=v(t)+c∫a(t)dt=v(t)+c ∫v(t)dx=s(t)+c∫v(t)dx=s(t)+c You need to find first the velocity function, such that:…

• MATH

a(t)=2t+1,s(0)=3,v(0)=−2a(t)=2t+1,s(0)=3,v(0)=-2 A particle is moving with the given data. Find the…

a(t)=2t+1a(t)=2t+1 a(t)=v'(t)a(t)=v′(t) v(t)=∫a(t)dtv(t)=∫a(t)dt v(t)=∫(2t+1)dtv(t)=∫(2t+1)dt v(t)=2(t22)+t+c1v(t)=2(t22)+t+c1 v(t)=t2+t+c1v(t)=t2+t+c1 Now let’s find constant c_1 given v(0)=-2 v(0)=−2=02+0+c1v(0)=-2=02+0+c1 c1=−2c1=-2 ∴v(t)=t2+t−2∴v(t)=t2+t-2…

• MATH

v(t)=1.5t√,s(4)=10v(t)=1.5t,s(4)=10 A particle is moving with the given data. Find the position…

As I understand we have to find s(t). s(t)=∫(v(t))dt=∫(1.5t√)dt=1.5⋅(23)⋅t32+C=t32+C,s(t)=∫(v(t))dt=∫(1.5t)dt=1.5⋅(23)⋅t32+C=t32+C,where C is any constant. Also is given that s(4)=10, so 432+C=10,432+C=10,…

• MATH

v(t)=sin(t)−cos(t),s(0)=0v(t)=sin(t)-cos(t),s(0)=0 A particle is moving with the given data. Find the…

v(t)=sin(t)−cos(t)v(t)=sin(t)-cos(t) v(t)=s'(t)v(t)=s′(t) s(t)=∫v(t)dts(t)=∫v(t)dt s(t)=∫(sin(t)−cos(t))dts(t)=∫(sin(t)-cos(t))dt s(t)=−cos(t)−sin(t)+Cs(t)=-cos(t)-sin(t)+C Let’s find the constant C , given s(0)=0 s(0)=0=−cos(0)−sin(0)+Cs(0)=0=-cos(0)-sin(0)+C0=−1−0+C0=-1-0+C C=1C=1 ∴∴…

• MATH

f(x)=x4−2×2+2−−−−−−−−−−√−2,−3≤x≤3f(x)=x4-2×2+2-2,-3≤x≤3 Draw a graph of ff and use it to make a…

Graph of ff can be seen in the picture below (blue line). From this we can sketch graph of its antiderivative. We start drawing from left to right. As long as ff is positive (graph above x-axis)…

• MATH

f(x)=sin(x)1+x2,−2π≤x≤2πf(x)=sin(x)1+x2,-2π≤x≤2π Draw a graph of ff and use it to make a…

Graph of ff can be seen in the picture below (blue line). From this we can sketch graph of its antiderivative. We start drawing from left to right. As long as ff is positive (graph above…

• MATH

f”'(x)=cos(x),f(0)=1,f'(0)=2,f”(0)=3f′′′(x)=cos(x),f(0)=1,f′(0)=2,f′′(0)=3 Find ff.

f”'(x)=cos(x)f′′′(x)=cos(x) f”(x)=∫cos(x)dxf′′(x)=∫cos(x)dx f”(x)=sin(x)+C1f′′(x)=sin(x)+C1Now let’s find C_1 , given f”(0)=3 f”(0)=3=sin(0)+C1f′′(0)=3=sin(0)+C1 3=0+C13=0+C1 C1=3C1=3 ∴f”(x)=sin(x)+3∴f′′(x)=sin(x)+3 f'(x)=∫(sin(x)+3)dxf′(x)=∫(sin(x)+3)dx…

• MATH

f”(x)=x−2,x>0,f(1)=0,f(2)=0f′′(x)=x-2,x>0,f(1)=0,f(2)=0 Find ff.

f”(x)=x−2f′′(x)=x-2 f'(x)=−x−1+af′(x)=-x-1+a f(x)=−lnx+ax+bf(x)=-lnx+ax+b Now,f(1)=f(2)=0Now,f(1)=f(2)=0 thus,a+b=0thus,a+b=0 and,−ln2+2a+b=0and,-ln2+2a+b=0 thus,a=ln2thus,a=ln2 b=−ln2b=-ln2 thus,f(x)=−lnx+xln2−ln2thus,f(x)=-lnx+xln2-ln2

• MATH

f”(t)=2et+3sin(t),f(0)=0,f(π)=0f′′(t)=2et+3sin(t),f(0)=0,f(π)=0 Find ff.

f”(t)=2et+3sin(t)f′′(t)=2et+3sin(t) f'(t)=∫(2et+3sin(t))dtf′(t)=∫(2et+3sin(t))dt f'(t)=2et−3cos(t)+C1f′(t)=2et-3cos(t)+C1 f(t)=∫(2et−3cos(t)+C1)dtf(t)=∫(2et-3cos(t)+C1)dt f(t)=2et−3sin(t)+C1t+C2f(t)=2et-3sin(t)+C1t+C2 Now let’s find constants C_1 and C_2 , given f(0)=0 and…

• MATH

f”(x)=2+cos(x),f(0)=−1,f(π2)=0f′′(x)=2+cos(x),f(0)=-1,f(π2)=0 Find ff.

f”(x)=2+cos(x)f′′(x)=2+cos(x) f'(x)=∫(2+cos(x)dxf′(x)=∫(2+cos(x)dx f'(x)=2x+sin(x)+C1f′(x)=2x+sin(x)+C1 f(x)=∫(2x+sin(x)+C1)dxf(x)=∫(2x+sin(x)+C1)dx f(x)=2(x22)−cos(x)+C1x+C2f(x)=2(x22)-cos(x)+C1x+C2 f(x)=x2−cos(x)+C1x+C2f(x)=x2-cos(x)+C1x+C2 Now lets find constants C_1 and C_2 , given…

• MATH

f”(x)=x3+sinh(x),f(0)=1,f(2)=2.6f′′(x)=x3+sinh(x),f(0)=1,f(2)=2.6 Find ff.

Integrate this once (all integrals are the table ones): f'(x)=(14)x4+cosh(x)+C1f′(x)=(14)x4+cosh(x)+C1 and twice: f(x)=(120)x5+sinh(x)+C1⋅x+C2.f(x)=(120)x5+sinh(x)+C1⋅x+C2. Now determine the constants C1C1 and C2C2 : f(0)=C2=1.f(0)=C2=1….

• MATH

f”(x)=4+6x+24×2,f(0)=3,f(1)=10f′′(x)=4+6x+24×2,f(0)=3,f(1)=10 Find ff.

f”(x)=4+6x+24x2f′′(x)=4+6x+24×2 f'(x)=∫(4+6x+24×2)dxf′(x)=∫(4+6x+24×2)dx f'(x)=4x+6×22+24×33+C1f′(x)=4x+6×22+24×33+C1 f'(x)=4x+3×2+8×3+C1f′(x)=4x+3×2+8×3+C1 f(x)=∫(4x+3×2+8×3+C1)dxf(x)=∫(4x+3×2+8×3+C1)dx f(x)=4×22+3×33+8×44+C1(x)+C2f(x)=4×22+3×33+8×44+C1(x)+C2…

• MATH

f”(t)=3t√,f(4)=20,f'(4)=7f′′(t)=3t,f(4)=20,f′(4)=7 Find ff.

f”(t)=3t√f′′(t)=3t f'(t)=∫(3t√)dtf′(t)=∫(3t)dt f'(t)=3(t−12+1−12+1)+C1f′(t)=3(t-12+1-12+1)+C1 f'(t)=6t√+C1f′(t)=6t+C1 Now let’s find constant C_1 given f'(4)=7, f'(4)=7=64–√+C1f′(4)=7=64+C1 7=12+C17=12+C1 C1=−5C1=-5…

• MATH

f”(θ)=sin(θ)+cos(θ),f(0)=3,f'(0)=4f′′(θ)=sin(θ)+cos(θ),f(0)=3,f′(0)=4 Find ff.

f”(θ)=sin(θ)+cos(θ)f′′(θ)=sin(θ)+cos(θ) f'(θ)=∫(sin(θ)+cos(θ))d(θ)f′(θ)=∫(sin(θ)+cos(θ))d(θ) f'(θ)=−cos(θ)+sin(θ)+C1f′(θ)=-cos(θ)+sin(θ)+C1 Now let’s find constant C_1 , given f'(0)=4 f'(0)=4=−cos(0)+sin(0)+C1f′(0)=4=-cos(0)+sin(0)+C1…

• MATH

f”(x)=8×3+5,f(1)=0,f'(1)=8f′′(x)=8×3+5,f(1)=0,f′(1)=8 Find ff.

f”(x)=8×3+5f′′(x)=8×3+5 f'(x)=∫(8×3+5)dxf′(x)=∫(8×3+5)dx f'(x)=8(x44)+5x+c1f′(x)=8(x44)+5x+c1 f'(x)=2×4+5x+c1f′(x)=2×4+5x+c1 Now let’s find constant c_1 , given f'(1)=8 f'(1)=8=2(1)4+5(1)+c1f′(1)=8=2(1)4+5(1)+c1 8=2+5+c18=2+5+c1 c1=1c1=1 ∴f'(x)=2×4+5x+1∴f′(x)=2×4+5x+1…

• MATH

f”(x)=−2+12x−12×2,f(0)=4,f'(0)=12f′′(x)=-2+12x-12×2,f(0)=4,f′(0)=12 Find ff.

f”(x)=−2+12x−12x2f′′(x)=-2+12x-12×2 f'(x)=∫(−2+12x−12×2)dxf′(x)=∫(-2+12x-12×2)dxf'(x)=−2x+12(x22)−12(x33)+c1f′(x)=-2x+12(x22)-12(x33)+c1 f'(x)=−2x+6×2−4×3+c1f′(x)=-2x+6×2-4×3+c1 Now let’s find constant c_1 , given f'(0)=12 f'(0)=12=−2(0)+6(02)−4(03)+c1f′(0)=12=-2(0)+6(02)-4(03)+c1…

• MATH

f'(x)=41−x2−−−−−√,f(12)=1f′(x)=41-x2,f(12)=1 Find ff.

f'(x)=41−x2−−−−−√f′(x)=41-x2 f(x)=∫(41−x2−−−−−√)dxf(x)=∫(41-x2)dx f(x)=4arcsin(x)+c1f(x)=4arcsin(x)+c1 Let’s find constant c_1 , given f(1/2)=1 f(12)=1=4arcsin(12)+c1f(12)=1=4arcsin(12)+c1 1=4(π6)+c11=4(π6)+c1 c1=1−2π3c1=1-2π3…

• MATH

f'(x)=x−13,f(1)=1f′(x)=x-13,f(1)=1 Find ff.

You need to evaluate f and the problem provides f'(x), hence, you need to use the following relation, such that: ∫f'(x)dx=f(x)+c∫f′(x)dx=f(x)+c You need to evaluate the indefinite integral of the power…

• MATH

f'(x)=x2−1x,f(1)=12,f(−1)=0f′(x)=x2-1x,f(1)=12,f(-1)=0 Find ff.

You need to evaluate the antiderivative of the function f'(x), such that: ∫f'(x)dx=f(x)+c∫f′(x)dx=f(x)+c ∫x2−1xdx=∫x2xdx−∫1xdx∫x2-1xdx=∫x2xdx-∫1xdx ∫x2−1xdx=∫xdx−∫1xdx∫x2-1xdx=∫xdx-∫1xdx

• MATH

f'(t)=2cos(t)+sec2(t),−π2<t<π2,f(π3)=4f′(t)=2cos(t)+sec2(t),-π2<t<π2,f(π3)=4 Find ff.

You need to evaluate the antiderivative of the function f'(t), such that: ∫f'(t)dt=f(t)+c∫f′(t)dt=f(t)+c ∫(2cost+sec2t)dt=∫2costdt+∫sec2tdt∫(2cost+sec2t)dt=∫2costdt+∫sec2tdt

• MATH

f'(t)=t+1t3,t>0,f(1)=6f′(t)=t+1t3,t>0,f(1)=6 Find ff.

Integrating gives f(t)=t22−(12)t−2+C,f(t)=t22-(12)t-2+C, where C is any constant. f(1)=12−12+C=C=6,f(1)=12-12+C=C=6, so C=6. The answer: f(t)=t22−12t2+6.f(t)=t22-12t2+6.

• MATH

f'(t)=41+t2,f(1)=0f′(t)=41+t2,f(1)=0 Find ff.

The general antiderivative of f’ is f(t)=4arctan(t)+C,f(t)=4arctan(t)+C, where C is any constant. It must be f(1)=0, so C=-4arctan(1). acrctan(1) is π4,π4, so the answer is f(t)=4arctan(t)−π.f(t)=4arctan(t)-π.