First slide

Introduction to integration

Question

If $f (x) d x = ϕ (x) i.e. ϕ$ is a function such that
$ϕ^{'} (x) = f (x),$ then $\int x^{9} f (x^{5}) d x$ is equal to

Moderate

A

$\frac{1}{5} [x^{5} ϕ (x^{5}) - \int x^{4} ϕ (x^{5}) d x] + C$
B

$\frac{1}{5} x^{5} ϕ (x^{5}) - 5 \int x^{5} ϕ (x^{5}) d x + C$
C

$\frac{1}{5} x^{5} ϕ (x^{5}) - \int x^{4} ϕ (x^{5}) d x + C$
D

$\frac{1}{5} [x^{5} ϕ (x^{5}) - \int x^{5} ϕ (x^{5}) d x] + C$

Solution

Let $x^{5} = t,$ so $\int x^{9} f (x^{5}) d x$
$\begin{array}{l} = \frac{1}{5} \int t f (t) d t \\ = \frac{1}{5} [t \int f (t) d t - \int ϕ (t) d t] \\ = \frac{1}{5} [x^{5} ϕ (x^{5}) - 5 \int ϕ (x^{5}) x^{4} d x] + C \\ = \frac{1}{5} x^{5} ϕ (x^{5}) - \int x^{4} ϕ (x^{5}) d x + C \end{array}$

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