substitution. Let u = e^3 x and du = 3e^3 x dx. Divide du by 3 so that du = e^3 x dx. Substituting
into the integrand, we get = tan−1 u + C. Substituting back, we get
tan−1 (e^3 x) + C.
15.
Remember that the derivative of sin−1 u = . Here, we get f′(θ) = .
Next, we plug in θ = : f′ = = . You could
simplify this some more to f′ , but it’s not necessary.
SOLUTIONS TO PRACTICE PROBLEM SET 24
1.
We find the area of a region bounded by f(x) above and g(x) below at all points of the interval
[a, b] using the formula. Here [f(x) − g(x)] dx. Here f(x) = 2 and g(x) = x^2 − 2. First, let’s
make a sketch of the region.
