Find the antiderivative?
Find the antiderivative of... A. f(x)=1/2 +(2/3)x^2-(6/7)x^3 B. f(x)=3x^12-2x^8+12x^4 C. f(x)=(x+3)(4x+3) D. f(x)=x(2-x)^2 E. f(x)=3x^(1/9)-8x^(8/9) F. f(x)=5x+6x^1.9 G. f(x)=2e^x+9sec^2x H. f(x)=cos(x)-2sin(x) I have solved the simplest ones ofcourse but I still dont fully understand how to work these. If you can help with any of these that would be amazing! Please answer and explanation!! Thanks so much!
How do I find the antiderivative of [math] \frac{4x^3-3x^2}{5x+7} [/math]?
Antiderivative of f(X) with....?
Take the antiderivative of f(x) (also known as an integral). This can be split up, the integral of 5/x^3 + integral of -7/x^7 5/x^3 can be rewritten as 5*x^(-3). Use the power rule. The antiderivative is -15*x^(-4), or -15/x^4 7/x^7 is the same process. Rewrite it as -7x^(-7), and use the power rule. The antiderivative is 49*x^-8, or 49/x^7 Adding the solutions, we get F(x) = -15/x^4 + 49/x^8 + C F(1)=0, so F(x)=0 when x = 1. 0=-15/(1)^4 + 49/(1)^8 + C This becomes: 0 = -15 + 49 + C Solving, C = -34 Finally, F(x) = -15/x^4 + 49/x^8 - 34
Find the most general antiderivative of the function.?
Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.) f(x) = 8x^9 − 4x^6 + 13x^3 F(x) = ??
Find the antiderivative of F of f(x)=4-3(1+x^2)^-1 that satisfies F(1)=-7?
F = ∫ [ 4 - 3(1 + x²)⁻¹ ] dx F = ∫ 4 dx - 3∫ [(1 + x²)⁻¹ ] dx F = 4x - 3arctan(x) + C -7 = 4(1) - 3arctan(1) + C -7 = 4 - 3π/4 + C C = -11 + 3π/4 Thus, F = 4x - 3arctan(x) -11 + 3π/4
How do you find the antiderivative of secxtanx?
Rewrite [math]secxtanx[/math][math]secxtanx=\frac{1}{cosx}\frac{sinx}{cosx}[/math][math]=\frac{sinx}{cos^2x}[/math]so, let find[math] \int\frac{sinx}{cos^2x}\,dx[/math]let [math]u = cosx[/math],[math]du=-sinx\,dx[/math] , and [math]dx=-\frac{1}{sinx}\,du[/math][math]\int\frac{sinx}{cos^2x}\,dx=\int\frac{sinx}{u^2}.-\frac{1}{sinx}\,du[/math][math]=-\int\frac{1}{u^2}\,du[/math][math]=-\int u^-2\,du[/math][math]=-\frac{1}{-2+1}u^{-2+1}[/math][math]=u^{-1}[/math][math]=\frac{1}{u}[/math][math]replace u=cosx [/math][math]\int secxtanx\,dx= \frac{1}{cosx}[/math][math]\int secxtanx\,dx=secx[/math]
Find an antiderivative of f(x) = tan^−1(x) satisfying F(0) = 3 using power series expansion.?
Given that,
Antiderivative help please?
I = (- 9) ∫ x^(- 4) dx I = (- 9) x^(- 3) / (- 3) + C I = 3 x^(- 3) + C I = 3 / x ³ + C