# Math Homework Help

### Showing All Questions Answered Popular Recommended Unanswered Editor's Choice in Math

• Math
We have to be careful here because of absolute value sign. We need to know where the function under absolute value changes sign (in the interval of integration), which is at x=pi in this case....

Asked by enotes on via web

• Math
Hello! Find the indefinite integral: int((sqrt(y)-y)/y^2)dy=int(y^(-3/2)-1/y)dy=(-2)*y^(-(1)/(2))-ln(y). Then substitute y from 1 to 4: (-2*4^(-1/2)-ln(4))-(-2)=-1-ln(4)+2=1-ln(4) approx -0.386.

Asked by enotes on via web

• Math
Hello! First find the indefinite integral, int(x/(2)-2/x)dx=int(x/2)dx-int(2/x)dx=(x^2)/4-2ln(x). Then substitute x from 1 to 2: (2^2/4-2ln(2))-(1^2/4-2ln(1))=1-2ln(2)-1/4=3/4-2ln(2) approx...

Asked by enotes on via web

• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_0^1(5x - 5^x)dx = int_0^1(5x)dx - int_0^1 5^x dx...

Asked by enotes on via web

• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_0^1(x^10 + 10^x)dx = int_0^1(x^10)dx + int_0^1 10^x...

Asked by enotes on via web

• 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_(pi/4)^(pi/3) csc^2 theta d theta = int_(pi/4)^(pi/3) 1/(sin^2...

Asked by enotes on via web

• 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_0^(pi/4)(1+cos^2 theta)/(cos^2 theta) d theta =...

Asked by enotes on via web

• Math
Before evaluating this integral, simplify the expression in the integral using trigonometric identities. The following Pythagorean identity will be useful: tan^2(theta) + 1 = sec^2(theta) Start...

Asked by enotes on via web

• Math
We will make substitution x=t^6. Therefore, the differential is dx=6t^5dt and the new bounds of integration are t_1=root(6)(1)=1 and t_2=root(6)(64)=2. int_1^64(1+root(3)(x))/sqrt x...

Asked by enotes on via web

• Math
Hello! Consider the denominator first: sinh(x)+cosh(x)=(e^x-e^(-x))/(2) +(e^x+e^(-x))/(2)=e^x. Therefore the integrand is equal to 2 and the integral is 2*(10-(-10))=40.

Asked by enotes on via web

• Math
Hello! This integral is a table one, int(dr)/(sqrt(1-r^2))=arcsin(r)+C.   Therefore the definite integral is equal to arcsin(sqrt(3)/2)-arcsin(0)=pi/3-0=pi/3 approx 1.047. =arcsin(r)+C.

Asked by enotes on via web

• 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) You need to expand the cube such that: (x-1)^3 = x^3 - 1 - 3x(x-1)...

Asked by enotes on via web

• Math
Hello! (t^2-1)/(t^4-1)=1/(t^2+1). Therefore indefinite integral is arctan(t)(+C). Substitute t from 0 to 1/sqrt(3) and obtain arctan(1/sqrt(3))-arctan(0)=pi/6-0=pi/6 approx 0.524.

Asked by enotes on via web

• Math
We have to be careful because of the absolute value. We need to know where the function under absolute values is positive and when it is negative in the given interval. It is also a good idea to...

Asked by enotes on via web

• Math
Hello! Consider two intervals: (-1, 0) and (0, 2) to simplify |x|: int_(-1)^2(x-2|x|)dx=int_(-1)^0(x-2|x|)dx+int_0^2(x-2|x|)dx= =int_(-1)^0(x+2x)dx+int_0^2(x-2x)dx=...

Asked by enotes on via web

• Math
You need to find the indefinite integral, hence, you need to remember that sec t = 1/(cos t) , such that: int (sec t)(sec t + tan t) dt = int (1/(cos t))(1 + sin t)/(cos t) dt int (sec t)(sec...

Asked by enotes on via web

• Math
You need to find the indefinite integral, hence, you need to remember that 1 + tan^2 x = 1/(cos^2 x) = (tan x)'. int (1 + tan^2 x )dx = int (tan x)' dx = tan x + c Hence, evaluating the...

Asked by enotes on via web

• Math
You need to find the indefinite integral, hence, you need to remember that sin 2x = 2sin x*cos x , such that: int (sin 2x)(sin x) dx = int (2sin x*cos x)/(sin x) dx  int (sin 2x)(sin x) dx =...

Asked by enotes on via web

• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_(-2)^3(x^3 - 3)dx = int_(-2)^3(x^3)dx - int_(-2)^3 3...

Asked by enotes on via web

• Math
You need to evaluate the integral, hence, you need to use the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a) int_1^2(4x^3 - 3x^2 + 2x)dx = int_1^2(4x^3)dx - int_1^2...

Asked by enotes on via web

• 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_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = int_(-2)^0 ((1/2)t^4 dt +...

Asked by enotes on via web

• Math
We will use linearity of integral: int(a cdot f(x)+b cdot g(x))dx=a int f(x)dx+b int g(x)dx. int_0^3(1+6w^2-10w^4)dw=int_0^3dw+6int_0^3w^2dw-10int_0^3w^4dw=...

Asked by enotes on via web

• Math
We have to evaluate the integral: \int_{0}^{2}(2x-3)(4x^2+1)dx=\int_{0}^{2}(8x^3-12x^2+2x-3)dx =[8(x^4/4)-12(x^3/3)+2(x^2/2)-3x]_{0}^{2}...

Asked by enotes on via web

• Math
You need to use the following substitution to evaluate the definite integral, such that: 1 - t = u => -dt = du int_(-1)^1 t*(1-t)^2dt = int_(u_1)^(u_2) (1 - u)*u^2 (-du) int_(u_1)^(u_2) (u...

Asked by enotes on via web

• 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_0^pi (5e^x+ 3sin x)dx = int_0^pi 5e^x dx + int_0^pi 3sin x...

Asked by enotes on via web

• 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_1^2 (1/x^2 - 4/x^3)dx = int_1^2 1/x^2 dx - int_1^2 4/x^3 dx...

Asked by enotes on via web

• Math
int_1^4(4+6u)/sqrt(u)du =int_1^4(4/sqrt(u)+(6u)/sqrt(u))du =int_1^4(4u^(-1/2)+6u^(1/2))du =[4(u^(-1/2+1)/(-1/2+1))+6(u^(1/2+1)/(1/2+1))]_1^4 =[4(u^(1/2)/(1/2))+6(u^(3/2)/(3/2))]_1^4...

Asked by enotes on via web

• 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_0^4(3sqrt t - 2e^t)dt = int_0^4 3sqrt tdt - int_0^4 2e^t...

Asked by enotes on via web

• Math
You need to evaluate the definite integral, such that: int_0^1 x(root(3) x + root(4) x)dx = int_0^1 (x^(1+1/3) + x^(1+1/4))dx int_0^1 x(root(3) x + root(4) x)dx = ((x^(2+1/3))/(2+1/3) +...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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, ...

Asked by enotes on via web

• 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 =...

Asked by enotes on via web

• 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

Asked by enotes on via web

• 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   

Asked by enotes on via web

• 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,...

Asked by enotes on via web

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

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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))....

Asked by enotes on via web

• 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)...

Asked by enotes on via web

• 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...

Asked by enotes on via web

• 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) =...

Asked by enotes on via web

• 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...

Asked by enotes on via web

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...