find the derivative using fundamental theorem of calculus integral of sin^3tdt from e^x to 0
d/dx ∫[e^x,0] sin^3(t) dt
= d/dx -∫[0,e^x] sin^3(t) dt
= -sin^3(e^x) * d/dx (e^x)
= -e^x sin^3(e^x)
To find the derivative using the Fundamental Theorem of Calculus, you need to follow these steps:
Step 1: Evaluate the integral.
Step 2: Apply the Fundamental Theorem of Calculus.
Step 3: Differentiate the resulting expression.
Let's solve the problem step by step:
Step 1: Evaluate the integral.
To begin, we need to find the integral of sin^3(t) dt from e^x to 0. To do this, we can use the standard integral formulas for trigonometric functions. The integral of sin^n(t) dt is given by:
∫ sin^n(t) dt = -1/n * cos(t) * sin^(n-1)(t) + (n-1)/n * ∫ sin^(n-2)(t) dt
Applying this formula, we can find the integral of sin^3(t) dt as follows:
∫ sin^3(t) dt = -1/3 * cos(t) * sin^2(t) + 2/3 * ∫ sin(t) dt
Now we have to find the integral of sin(t) dt. Differentiating -cos(t) will give us sin(t), so we can rewrite the integral as:
∫ sin(t) dt = -cos(t)
Now we can substitute this result back into the original integral:
∫ sin^3(t) dt = -1/3 * cos(t) * sin^2(t) + 2/3 * (-cos(t))
Step 2: Apply the Fundamental Theorem of Calculus.
According to the Fundamental Theorem of Calculus, if we have an integral from a to b (in this case, from e^x to 0), the derivative with respect to x is given by:
d/dx ∫ f(t) dt = f(b) * d/dx b - f(a) * d/dx a
In our case, f(t) = ∫ sin^3(t) dt, a = e^x, and b = 0. Therefore, we have:
d/dx ∫ sin^3(t) dt = ∫ sin^3(0) dt - ∫ sin^3(e^x) dt
Step 3: Differentiate the resulting expression.
We can simplify this further. sin(0) = 0, so the first term on the right-hand side becomes zero:
d/dx ∫ sin^3(t) dt = - ∫ sin^3(e^x) dt
And now we have the derivative of the original integral using the Fundamental Theorem of Calculus. The resulting expression is - ∫ sin^3(e^x) dt.