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…
1 educator answer
- MATH
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…
1 educator answer
- 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
1 educator answer
- 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:
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- 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.
1 educator answer
- 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…
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- SOCIAL SCIENCES
Can someone please help me with the following Global Policy – American Government questions….
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….
1 educator answer
- MATH
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…
1 educator answer
- MATH
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…
1 educator answer
- MATH
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:…
1 educator answer
- 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…
1 educator answer
- 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,…
1 educator answer
- MATH
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 ∴∴…
1 educator answer
- 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)…
1 educator answer
- 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…
2 educator answers
- 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…
1 educator answer
- 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
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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….
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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…
1 educator answer
- 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
1 educator answer
- 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
1 educator answer
- 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.
1 educator answer
- 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.
1 educator answer
- MATH
f'(x)=5×4−3×2+4,f(−1)=2f′(x)=5×4-3×2+4,f(-1)=2 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 ∫(5×4−3×2+4)dx=f(x)+c∫(5×4-3×2+4)dx=f(x)+c You need to evaluate…
1 educator answer