Question

(a) evaluate the indefinite integral $\int(3 x^{2}-2 x + 10) d x$.
(b) evaluate the definite integral $\int_{0}^{1}(3 x^{2}-2 x + 10) d x$
(a) evaluate the integral.
$\int(3 x^{2}-2 x + 10) d x=6\left(\frac{x^{2}}{2}-2 x\
ight)+c$

Explanation:

Step1: Integrate each term separately

Use the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For \(\int3x^{2}dx\), \(n = 2\), so \(\int3x^{2}dx=3\times\frac{x^{2+1}}{2 + 1}=x^{3}\).
For \(\int(-2x)dx\), \(n = 1\), so \(\int(-2x)dx=-2\times\frac{x^{1+1}}{1+1}=-x^{2}\).
For \(\int10dx\), since \(\int kdx=kx + C\) (\(k\) is a constant), \(\int10dx=10x\).

Step2: Combine the results

\(\int(3x^{2}-2x + 10)dx=\int3x^{2}dx-\int2xdx+\int10dx=x^{3}-x^{2}+10x + C\)

Step3: Evaluate the definite integral

Use the fundamental theorem of calculus \(\int_{a}^{b}f(x)dx=F(b)-F(a)\), where \(F(x)\) is the antiderivative of \(f(x)\).
We know \(F(x)=x^{3}-x^{2}+10x\), \(a = 0\), \(b = 1\).
\(F(1)=1^{3}-1^{2}+10\times1=1 - 1+10 = 10\).
\(F(0)=0^{3}-0^{2}+10\times0=0\).
\(\int_{0}^{1}(3x^{2}-2x + 10)dx=F(1)-F(0)=10-0 = 10\)

Answer:

(a) \(\int(3x^{2}-2x + 10)dx=x^{3}-x^{2}+10x + C\)
(b) \(\int_{0}^{1}(3x^{2}-2x + 10)dx = 10\)