Question

Please help me i swear i’ll give u crown. INTEGRALS INTEGRALS

the other part is uploaded in my profile INTEGRALS PLEASE HELP ME

Answer 1

`\textsf{1)} \quad 4.1875`

`\textsf{2)} \quad (1)/(2)x^(4)-x^(3)+x^(2)-3x+\text{C}`

Explanation:

Fundamental Theorem of Calculus

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration:

`\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\frac{\text{d}}{\text{d}x}(\text{F}(x))`

The area between a curve and the x-axis can be found using definite integration. The definite integral of f(x) with respect to x between the limits x = a and x = b:

`\displaystyle \int^b_a \text{f}(x)\; \text{d}x=\left[\text{g}(x)\right]^b_a=\text{g}(b)-\text{g}(a)`

where a is the lower limit and b is the upper limit.

`\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=(x^(n+1))/(n+1)\left(+\text{C}\right)$\end{minipage}}`

Question 1

Given function:

`\text{f}(x)=(1)/(4)x^3+(1)/(6)x+1`

From inspection of the given graph:

• Lower limit a = -1
• Upper limit b = 2

Therefore:

`\begin{aligned}\displaystyle\int^2_(-1)\left((1)/(4)x^3+(1)/(6)x+1\right)\;\text{d}x&=\left[(1)/(4\cdot4)x^(3+1)+(1)/(6\cdot 2)x^(1+1)+x\right]^2_(-1)\\\\&=\left[(1)/(16)x^4+(1)/(12)x^2+x\right]^2_(-1)\\\\&=\left((1)/(16)(2)^(4)+(1)/(12)(2)^2+(2) \right)-\left((1)/(16)(-1)^4+(1)/(12)(-1)^2+(-1)\right)\\\\&=\left(1+(1)/(3)+2\right)-\left((1)/(16)+(1)/(12)-1\right)\\\\&=(10)/(3)+(41)/(48)\\\\&=(67)/(16)\\\\&=4.1875\end{aligned}`

Question 2

`\begin{aligned}\displaystyle \displaystyle \int (x^2+1)(2x-3)\; \text{d}x&=\displaystyle \int \left(2x^3-3x^2+2x-3\right)\; \text{d}x\\\\&=(2)/(4)x^(3+1)-(3)/(3)x^(2+1)+(2)/(2)x^(1+1)-3x+\text{C}\\\\&=(1)/(2)x^(4)-x^(3)+x^(2)-3x+\text{C}\end{aligned}`