Question

evaluate the following integral using the fundamental theorem of calculus.
int_{0}^{10} t(t - 2)(t - 10) dt
int_{0}^{10} t(t - 2)(t - 10) dt=square (type an exact answer.)

Explanation:

Step1: Expand the integrand

First, we expand \( t(t - 2)(t - 10) \).
First multiply \( (t - 2)(t - 10) = t^2-10t - 2t + 20=t^2-12t + 20 \).
Then multiply by \( t \): \( t(t^2-12t + 20)=t^3-12t^2 + 20t \).
So the integral becomes \( \int_{0}^{10}(t^3-12t^2 + 20t)dt \).

Step2: Integrate term - by - term

Using the power rule for integration \( \int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)):
\(\int(t^3-12t^2 + 20t)dt=\frac{t^{4}}{4}-12\times\frac{t^{3}}{3}+20\times\frac{t^{2}}{2}+C=\frac{t^{4}}{4}-4t^{3}+10t^{2}+C\)

Step3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that \( \int_{a}^{b}f(x)dx=F(b)-F(a) \), where \( F(x) \) is an antiderivative of \( f(x) \).
Here, \( a = 0 \), \( b = 10 \), and \( F(t)=\frac{t^{4}}{4}-4t^{3}+10t^{2} \).
Evaluate \( F(10) \):
\(F(10)=\frac{10^{4}}{4}-4\times10^{3}+10\times10^{2}=\frac{10000}{4}-4000 + 1000=2500-4000 + 1000=- 500\)
Evaluate \( F(0) \):
\(F(0)=\frac{0^{4}}{4}-4\times0^{3}+10\times0^{2}=0\)
Then \( \int_{0}^{10}(t^3-12t^2 + 20t)dt=F(10)-F(0)=-500 - 0=-500\)

Answer:

\(-500\)