# Homework Help

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• Math
You need to evaluate the indefinite integral, such that: int f(x)dx = F(x) + c int (x^2 - x^(-2))dx = int (x^2)dx - int x^(-2) dx  Evaluating each definite integral, using the formula int x^n...

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• Math
int sqrt(x^3)root(3)(x^2) dx Before evaluating, convert the radicals to expressions with rational exponents. = int x^(3/2)*x^(2/3) dx Then, simplify the integrand. Apply the laws of exponent...

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• Math
int (x^4-1/2x^3+1/4x-2)dx To evaluate this integral, apply the formulas int x^n dx=x^(n+1)/(n+1) +C and int adx = ax + C . int (x^4-1/2x^3+1/4x-2)dx =x^5/5 - 1/2*x^4/4 + 1/4*x^2/2-2x...

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• Math
int (y^3+1.8y^2-2.4y)dy To evaluate this integral, apply the formula int x^n dx = x^(n+1)/(n+1) + C . = y^4/4 + 1.8y^3/3 - 2.4y^2/2 + C =0.25y^4 + 0.6y^3 - 1.2y^2 + C Therefore, ...

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• Math
You need to evaluate the indefinite integral, hence, you need to open the brackets, such that: (u+4)(2u+1) = 2u^2 + 9u + 4 int (u+4)(2u+1) du = int (2u^2 + 9u + 4) du  int (u+4)(2u+1) du =...

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• Math
intv(v^2+2)^2dv =intv((v^2)^2+2v^2*2+2^2)dv =intv(v^4+4v^2+4)dv =int(v^5+4v^3+4v)dv apply the power rule, =v^6/6+4v^4/4+4v^2/2+C , C is constant =v^6/6+v^4+2v^2+C

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• Math
int(x^3-2sqrt(x))/xdx Simplify by dividing each term in the numerator by x.  <br>  =(x^3/3)-(2x^(1/2))/(1/2)+C =(x^3/3)-4x^(1/2)+C  

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• Math
int(x^2+1+1/(x^2+1))dx apply the sum rule, =intx^2dx+int1dx+int1/(x^2+1)dx To evaluate the above integrals, we know that, intx^ndx=x^(n+1)/(n+1) and int1/(x^2+1)dx=arctan(x) using above,...

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• Math
We have to find the integral : \int (sin(x)+sinh(x) )dx=-cos(x)+cosh(x)+C where C is a constant.

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• Math
int(csc^2(t)-2e^t)dt Apply the sum rule, =intcsc^2(t)dt-int(2e^tdt We now the following common integrals, intcsc^2(x)=-cot(x) and inte^xdx=e^x evaluate using the above, =-cot(t)-2e^t+C...

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• Math
You need to evaluate the indefinite integral, such that: int f(theta)d theta = F(theta) + c int (theta - csc theta* cot theta)d theta = int theta d theta - int (csc theta* cot theta)d theta...

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• Math
y=sec^2(x) Refer the graph in the attached image. From the graph, Area of the region beneath the curve ~~ 4/10(Area of the Rectangle) Area of the region=~~(4/10)(pi/3*4) Area of the...

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• Math
We have to evaluate the integral and interpret as difference of areas. So, \int_{-1}^{2}x^3dx=[x^4/4]_{-1}^{2} =[(2^4)/4]-1/4 =16/4-1/4 =15/4=3.75 So...

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• Math
The integral actually represents difference between red area and green area shown in the image below i.e. int_(pi/6)^(2pi)cos(x)dx=P_1-P_2+P_3...

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• Math
g(x)=int_(2x)^(3x)(u^2-1)/(u^2+1)du g(x)=int_(2x)^0(u^2-1)/(u^2+1)du+int_0^(3x)(u^2-1)/(u^2+1)du g(x)=-int_0^(2x)(u^2-1)/(u^2+1)du+int_0^(3x)(u^2-1)/(u^2+1)du...

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• Math
Hello! Let's temporarily denote the antiderivative of tsin(t) as A(t). Then the integral is equal to A(1+2x)-A(1-2x) and its derivative is A'(1+2x)*2-A'(1-2x)*(-2)=2*(A'(1+2x)+A'(1-2x))....

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• Math
F(x)=int_x^(x^2)(e^(t^2))dt From the fundamental theorem of calculus, int_x^(x^2)(e^(t^2))dt=F(x^2)-F(x) d/dxint_x^(x^2)(e^(t^2))dt=F'(x^2).d/dx(x^2)-F'(x) =2x(e^((x^2)^2))-e^(x^2)...

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• Math
Hello! Let's temporarily denote the antiderivative of arctan(t) as A(t). Then F(x)= A(2x)-A(sqrt(x)), and F'(x)=A'(2x)*2-A'(sqrt(x))*(1/(2sqrt(x))). Recall what A'(x) is and obtain...

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• Math
You need to evaluate the the derivative of the function, hence, you need to use the part1 of fundamental theorem of calculus: y = int_a^b f(x)dx => (dy)/(dx) = f(x) for x in (a,b) If f(x) =...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(u) du = F(b) - F(a) int_0^3 (2sin x - e^x) dx =int_0^3 2sin x dx - int_0^3 e^x dx...

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• Math
int_1^2((v^3+3v^6)/v^4)dv simplify the integrand and apply the sum rule, =int_1^2(v^3/v^4+(3v^6)/v^4)dv =int_1^2(1/v+3v^2)dv using the following common integrals int1/xdx=ln(x) and...

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• Math
Evaluate int_1^18(3/z)^(1/2)dz =int_1^18sqrt(3)z^(-1/2)dz =sqrt(3)int_1^18z^(-1/2)dz Integrate the function. =sqrt(3)[z^(1/2)/(1/2)]=sqrt(3)[2z^(1/2)] Evaluate the function from 1 to 18....

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• Math
int_0^1 (x^e+e^x)dx To evaluate this, apply the formulas int u^ndu = u^(n+1)/(n+1) and int e^udu=e^u . = (x^(e+1)/(e+1) + e^x) |_0^1 Then, plug-in the limits of integral as follows...

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• Math
We have to evaluate the integral \int_{0}^{1}cosh(t)dt  We know that the integral of cosh(t) = sinh(t) . Therefore we can write, \int_{0}^{1}cosh(t)dt=[sinh(t)]_{0}^{1}...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x) dx = F(b) - F(a) int_(1/(sqrt3))^(sqrt 3) 8/(1+x^2) dx = 8 int_(1/(sqrt3))^(sqrt 3)...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(u) du = F(b) - F(a) int_1^2 (4+u^2)/(u^3) du = int_1^2 4/(u^3) du + int_1^2 (u^2)/(u^3)...

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• Math
int_(-1)^1 e^(u+1)du To evaluate this, apply the formula int e^x dx = e^x . = e^(u+1) |_(-1)^1 Then, plug-in the limits of the integral as follows F(x) =int_a^b f(x)dx=F(b)-F(a) ....

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• Math
Hello! This integral is a table one, int((4)/(sqrt(1-x^2)))dx=4arcsin(x)+C. Therefore the definite integral is equal to 4*(arcsin(1/sqrt(2))-arcsin(1/2))=4*(pi/4-pi/6)=4*pi/12=pi/3 approx 1.047.

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• Math
Refer the graph in the attached image. From the graph it appears that area of the region is ~~ 70% of the rectangle. Area of the region=~~ 70/100(Area of Rectangle) Area of the region...

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• Math
Refer the graph in the attached image. From the graph, Area of the region that lies beneath the curve ~~ 6/100(Area of Rectangle) Area of the region ~~ 6/100(5*1) ~~ 0.30 Exact Area of the...

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• Math
Refer the graph in the attached image. From the graph it appears that the area is ~~ 2/3 of the rectangle. Area of the region= ~~ 2/3(Area of rectangle) Area of the region...

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• Math
You need to evaluate the definite integral using the fundamental theorem of calculus such that int_a^b f(x)dx = F(b) - F(a) int_(pi/6)^pi sin theta d theta = -cos theta|_(pi/6)^pi...

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• Math
Evaluate int_-5^5(e)dx Please note that e is a constant approximately equal to 2.718. Integrate the function. =ex Evaluate the function from x=-5 to x=5. =e(5)-e(-5)=5e+5e=10e =27.183

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• Math
int_0^1(u+2)(u-3)du =int_0^1(u^2-3u+2u-6)du =int_0^1(u^2-u-6)du =[u^3/3-u^2/2-6u]_0^1 =[1^3/3-1^2/2-6*1]-[0^3/3-0^2/2-6*0] =(1/3-1/2-6) =(2-3-36)/6 =-37/6

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• Math
Evaluate int_0^4(4-t)(sqrt(t))dt =int_0^4(4t^(1/2)-t^(3/2))dt Integrate the function. inta^n=a^(n+1)/(n+1) =(4t^(3/2))/(3/2)-t^(5/2)/(5/2)=(8/3)t^(3/2)-(2/5)t^(5/2) Evaluate the function...

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• Math
Hello! Find the indefinite integral first: int((x-1)/sqrt(x))dx=int(x^(1/2)-x^(-1/2))dx=(2/3)*x^(3/2)-2*x^(1/2)+C. So the definite integral is equal to...

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• Math
int_0^2 (y-1)(2y+1)dy Before evaluating, expand the integrand. =int_0^2 (2y^2+y-2y-1)dy =int_0^2(2y^2-y-1)dy Then, apply the integral formulas int x^n dx=x^(n+1)/(n+1) and int cdx = cx ....

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• Math
int_0^(pi/4) sec^2(t) dt Take note that the derivative of tangent is d/(d theta) tan (theta)= sec^2 (theta). So taking the integral of sec^2(t) result to: = tan (t) |_0^(pi/4) Plug-in the...

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• Math
int_0^(pi/4) (sec (theta) tan (theta)) d theta Take note that the derivative of secant is d/(d theta) (sec (theta)) = sec(theta) tan (theta) . So taking the integral of sec(theta) tan(theta)...

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• Math
int_1^2(1+2y)^2dy =int_1^2((1)^2+2*2y*1+(2y)^2)dy =int_1^2(1+4y+4y^2)dy =[y+4y^2/2+4y^3/3]_1^2 =[y+2y^2+(4y^3)/3]_1^2 =[2+2(2)^2+(4(2^3))/3]-[1+2(1)^2+(4(1)^3)/3]...

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• Math
int_1^4(5-2t+3t^2)dt apply the sum rule and power rule, =[5t-2t^2/2+3t^3/3]_1^4 =[5t-t^2+t^3]_1^4 =[5*4-4^2+4^3]-[5*1-1^2+1^3] =(20-16+64)-(5-1+1) =(84-16)-(5) =63

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• Math
int_0^1(1+(1/2)u^4+(2/5)u^9)du =[u+(1/2)(u^(4+1)/(4+1))+(2/5)(u^(9+1)/(9+1))]_0^1 =[u+u^5/10-u^10/25]_0^1 = [1+1^5/10-1^10/25]-[0+0^5/10-0^10/25] =(1+1/10-1/25) =(50+5-2)/50 =53/50

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• Math
You need to evaluate the definite integral such that: int_1^9 sqrt x dx = (x^(3/2))/(3/2)|_1^9 int_1^9 sqrt x dx = (2/3)(9sqrt9 - 1sqrt1) int_1^9 sqrt x dx = (2/3)(27-1) int_1^9 sqrt x dx...

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• Math
Evaluate int_1^8(x^(-2/3))dx Integrate the function. inta^n=a^(n+1)/(n+1) =x^(1/3)/(1/3)=3x^(1/3) Evaluate the function from x=1 to x=8. =3[8^(1/3)-1^(1/3)] =3[2-1] =3

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• Math
Hello! Part 1 of the Fundamental Theorem of Calculus states that for a continuous function f F'_a(x)=f(x), where F_a(x)=int_a^xf(t)dt. Here f(t)=sqrt(t^2+4) and g(x)=F_0(x)....

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• Math
F(x)=int_x^pisqrt(1+sec(t))dt F(x)=-int_pi^xsqrt(1+sec(t))dt F'(x)=-d/dxint_pi^xsqrt(1+sec(t))dt F'(x)=-sqrt(1+sec(x))

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• Math
You need to use the Part 1 of the FTC to evaluate the derivative of the function. You need to notice that the function G(x) is the composite of two functions f(x) = int_1^x cos t dt  and g(x) =...

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• Math
Hello! Part 1 of the Fundamental Theorem of Calculus states that for a continuous function f F'_a(x)=f(x), where F_a(x)=int_a^xf(t)dt. Here f(t)=ln(t) and h(x)=F_1(e^x). Therefore...

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• Math
Hello! Part 1 of the Fundamental Theorem of Calculus states that for a continuous function f F'_a(x)=f(x), where F_a(x)=int_a^xf(t)dt. Here f(t)=t^2/(t^4+1) and h(x)=F_1(sqrt(x))...

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