0
Add your answer:
Earn +20 pts
What is a number to the power of 10?
To get the power of a number you multiply it by itself the specified number of times. For instance, the power 2 to the power of 3 would be 2 X 2 X 2 w 2 which would be 8. The power of 2 to power of 10 would therefore be 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 which would equal 1024. 10 to the power of 10 would be 10,000,000,000. 1 to the power of 10 would be 1.
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
What is the integral of x divided by sq-rt of x plus 2?
Integral of x dx / sqrt(x+2) Make the substitution sqrt(x+2)=u (x+2)^(1/2) = u (1/2)(x+2)^(-1/2) dx = du 1/2(x+2)^(1/2) dx = du 1/2sqrt(x+2) dx = du 1/sqrt(x+2) dx = 2 du Integral of x dx / sqrt(x+2) = Integral 2 x du sqrt(x+2) = u (x+2)=u^2 x=u^2-2 Integral 2 x du = Integral 2(u^2-2) du = Integral 2u^2 du - 4 du = 2 u^3/3 - 4u + C = (2/3) (x+2)^(3/2) - 4 sqrt(x+2) + C
What is the integral of 3 over 3x?
The 3s would cancel and it would become the integral of 1/x which is ln x.
How do you find the integrals of x?
You add one to the power, and then divide by the power that it has so you would have: Integral of x = (x^2)/2 Integral of x^2 = (x^3)/3 Etc.
Integral of e to the power of -x?
integral of e to the power -x is -e to the power -x
How do you evaluate definite integrals?
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
What is the integral of dx divided by x-1 to the power of 2?
-(x-1)-1 or -1/(x-1)
Integration of tan pow4x?
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
What is the integral of 2-x?
The integral of 2-x = 2x - (1/2)x2 + C.
What is the integral of 2 times x divided by the quantity x plus 1 to the third power?
3
What is a number to the power of 10?
To get the power of a number you multiply it by itself the specified number of times. For instance, the power 2 to the power of 3 would be 2 X 2 X 2 w 2 which would be 8. The power of 2 to power of 10 would therefore be 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 which would equal 1024. 10 to the power of 10 would be 10,000,000,000. 1 to the power of 10 would be 1.
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
Integral of e to the power of x?
The antiderivative, or indefinite integral, of ex, is ex + C.
What is the antiderivative of -1x 1?
If the term is -x, the integral expression is simply -∫x. By undoing the power rule, we get -(1/2)x^2+C, an arbitrary constant.
